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Chemistry LibreTexts

5.1.1: Practice Problems- Writing and Balancing Chemical Equations

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PROBLEM \(\PageIndex{1}\)

Balance the following equations:

  • \(\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{HCl}(aq)\)
  • \(\ce{Cu}(s)+\ce{HNO3}(aq)\rightarrow \ce{Cu(NO3)2}(aq)+\ce{H2O}(l)+\ce{NO}(g)\)
  • \(\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{HI}(s)\)
  • \(\ce{Fe}(s)+\ce{O2}(g)\rightarrow \ce{Fe2O3}(s)\)
  • \(\ce{Na}(s)+\ce{H2O}(l)\rightarrow \ce{NaOH}(aq)+\ce{H2}(g)\)
  • \(\ce{(NH4)2Cr2O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{H2O}(g)\)
  • \(\ce{P4}(s)+\ce{Cl2}(g)\rightarrow \ce{PCl3}(l)\)
  • \(\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{Cl2}(g)\)

\(\ce{PCl5}(s)+\ce{H2O}(l)\rightarrow \ce{POCl3}(l)+\ce{2HCl}(aq)\)

\(\ce{3Cu}(s)+\ce{8HNO3}(aq)\rightarrow \ce{3Cu(NO3)2}(aq)+\ce{4H2O}(l)+\ce{2NO}(g)\)

\(\ce{H2}(g)+\ce{I2}(s)\rightarrow \ce{2HI}(s)\)

\(\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)\)

\(\ce{2Na}(s)+\ce{2H2O}(l)\rightarrow \ce{2NaOH}(aq)+\ce{H2}(g)\)

\(\ce{(NH4)2Cr52O7}(s)\rightarrow \ce{Cr2O3}(s)+\ce{N2}(g)+\ce{4H2O}(g)\)

\(\ce{P4}(s)+\ce{6Cl2}(g)\rightarrow \ce{4PCl3}(l)\)

\(\ce{PtCl4}(s)\rightarrow \ce{Pt}(s)+\ce{2Cl2}(g)\)

PROBLEM \(\PageIndex{2}\)

  • \(\ce{Ag}(s)+\ce{H2S}(g)+\ce{O2}(g)\rightarrow \ce{Ag2S}(s)+\ce{H2O}(l)\)
  • \(\ce{P4}(s)+\ce{O2}(g)\rightarrow \ce{P4O10}(s)\)
  • \(\ce{Pb}(s)+\ce{H2O}(l)+\ce{O2}(g)\rightarrow \ce{Pb(OH)2}(s)\)
  • \(\ce{Fe}(s)+\ce{H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{H2}(g)\)
  • \(\ce{Sc2O3}(s)+\ce{SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)\)
  • \(\ce{Ca3(PO4)2}(aq)+\ce{H3PO4}(aq)\rightarrow \ce{Ca(H2PO4)2}(aq)\)
  • \(\ce{Al}(s)+\ce{H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{H2}(g)\)
  • \(\ce{TiCl4}(s)+\ce{H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{HCl}(g)\)

\(\ce{4Ag}(s)+\ce{2H2S}(g)+\ce{O2}(g)\rightarrow \ce{2Ag2S}(s)+\ce{2H2O}(l)\)

\(\ce{P4}(s)+\ce{5O2}(g)\rightarrow \ce{P4O10}(s)\)

\(\ce{2Pb}(s)+\ce{2H2O}(l)+\ce{O2}(g)\rightarrow \ce{2Pb(OH)2}(s)\)

\(\ce{3Fe}(s)+\ce{4H2O}(l)\rightarrow \ce{Fe3O4}(s)+\ce{4H2}(g)\)

\(\ce{Sc2O3}(s)+\ce{3SO3}(l)\rightarrow \ce{Sc2(SO4)3}(s)\)

\(\ce{Ca3(PO4)2}(aq)+\ce{4H3PO4}(aq)\rightarrow \ce{3Ca(H2PO4)2}(aq)\)

\(\ce{2Al}(s)+\ce{3H2SO4}(aq)\rightarrow \ce{Al2(SO4)3}(s)+\ce{3H2}(g)\)

\(\ce{TiCl4}(s)+\ce{2H2O}(g)\rightarrow \ce{TiO2}(s)+\ce{4HCl}(g)\)

PROBLEM \(\PageIndex{3}\)

Write a balanced molecular equation describing each of the following chemical reactions.

  • Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
  • Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.
  • Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
  • Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.

\(\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(g)\)

\(\ce{2C4H10}(g)+\ce{13O2}(g)\rightarrow \ce{8CO2}(g)+\ce{10H2O}(g)\)

\(\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)\)

\(\ce{2H2O}(g)+\ce{2Na}(s)\rightarrow \ce{2NaOH}(s)+\ce{H2}(g)\)

PROBLEM \(\PageIndex{4}\)

Write a balanced equation describing each of the following chemical reactions.

  • Solid potassium chlorate, KClO 3 , decomposes to form solid potassium chloride and diatomic oxygen gas.
  • Solid aluminum metal reacts with solid diatomic iodine to form solid Al 2 I 6 .
  • When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.
  • Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.

\(\ce{2KClO3}(s)\rightarrow \ce{2KCl}(s)+\ce{3O2}(g)\)

\(\ce{2Al}(s)+\ce{3I2}(s)\rightarrow \ce{Al2I6}(s)\)

\(\ce{2NaCl}(s)+\ce{H2SO4}(aq)\rightarrow \ce{2HCl}(g)+\ce{Na2SO4}(aq)\)

\(\ce{H3PO4}(aq)+\ce{KOH}(aq)\rightarrow \ce{KH2PO4}(aq)+\ce{H2O}(l)\)

PROBLEM \(\PageIndex{5}\)

Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.

  • Write the formulas of barium nitrate and potassium chlorate.
  • The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.
  • The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.
  • Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe 3+ ions.)

Ba(NO 3 ) 2 , KClO 3

\(\ce{2Ba(NO3)2}(s)\rightarrow \ce{2BaO}(s)+\ce{2N2}(g)+\ce{5O2}(g)\)

\(\ce{2Mg}(s)+\ce{O2}(g)\rightarrow \ce{2MgO}(s)\) ; \(\ce{4Al}(s)+\ce{3O2}(g)\rightarrow \ce{2Al2O3}(g)\); \(\ce{4Fe}(s)+\ce{3O2}(g)\rightarrow \ce{2Fe2O3}(s)\)

PROBLEM \(\PageIndex{6}\)

Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).

  • Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.
  • The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write the equation for this reaction.

\(\ce{4HF}(aq)+\ce{SiO2}(s)\rightarrow \ce{SiF4}(g)+\ce{2H2O}(l)\)

\(\ce{CaCl2}(aq)+\ce{2NaF}(aq)\rightarrow \ce{2NaCl}(aq)+\ce{CaF2}(s)\)

PROBLEM \(\PageIndex{7}\)

A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.

  • The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.
  • The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.
  • Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.
  • The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.
  • Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.

\(\ce{CaO}(s)+\ce{H2O}(l)\rightarrow \ce{Ca(OH)2}(s)\)

\(\ce{Ca(OH)2}(s)+\ce{MgCl2}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{CaCl2}(aq)\)

\(\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{2H2O}(l)\)

\(\ce{MgCl2}(s)\rightarrow \ce{Mg}(s)+\ce{Cl2}(g)\)

Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.  Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).

  • Adelaide Clark, Oregon Institute of Technology

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Balancing Chemical Equations Practice Sheet

Balancing chemical equations requires practice. Once you’ve done it a few times, it becomes easier and easier. This balancing chemical equations practice sheet has ten more unbalanced chemical equations to solve.

Balance Chemical Equations Worksheet #2

Download a PDF of this worksheet here .

A PDF of the answer key is also available here . If you’d just like to check your answers, click here to see the completed sheet.

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This online quiz is intended to give you extra practice in balancing, identifying and predicting a random selection of over 150 chemical equations. This quiz aligns with the following NGSS standard(s): HS-PS1-2 , HS-PS1-7

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practice problems of balancing chemical equations

Balancing Equations Test Questions

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Chemical reactions have the same number of atoms before the reaction as after the reaction. Balancing chemical equations is a basic skill in chemistry and testing yourself helps retain important information. This collection of ten chemistry test questions will give you practice in how to balance chemical reactions .

Balance the following equation:

__ SnO 2 + __ H 2 → __ Sn + __ H 2 O

__ KOH + __ H 3 PO 4 → __ K 3 PO 4 + __ H 2 O

__ KNO 3 + __ H 2 CO 3 → __ K 2 CO 3 + __ HNO 3

Balance the following equation: __ Na 3 PO 4 + __ HCl → __ NaCl + __ H 3 PO 4

__ TiCl 4 + __ H 2 O → __ TiO 2 + __ HCl

__ C 2 H 6 O + __ O 2 → __ CO 2 + __ H 2 O

__ Fe + __ HC 2 H 3 O 2 → __ Fe(C 2 H 3 O 2 ) 3 + __ H 2

__ NH 3 + __ O 2 → __ NO + __ H 2 O

__ B 2 Br 6 + __ HNO 3 → __ B(NO 3 ) 3 + __ HBr

Question 10

__ NH 4 OH + __ Kal(SO 4 ) 2 ·12H 2 O → __ Al(OH) 3 + __ (NH 4 ) 2 SO 4 + __ KOH + __ H 2 O

1. 1 SnO 2 + 2 H 2 → 1 Sn + 2 H 2 O 2. 3 KOH + 1 H 3 PO 4 → 1 K 3 PO 4 + 3 H 2 O 3. 2 KNO 3 + 1 H 2 CO 3 → 1 K 2 CO 3 + 2 HNO 3 4. 1 Na 3 PO 4 + 3 HCl → 3 NaCl + 1 H 3 PO 4 5. 1 TiCl 4 + 2 H 2 O → 1 TiO 2 + 4 HCl 6. 1 C 2 H 6 O + 3 O 2 → 2 CO 2 + 3 H 2 O 7. 2 Fe + 6 HC 2 H 3 O 2 → 2 Fe(C 2 H 3 O 2 ) 3 + 3 H 2 8. 4 NH 3 + 5 O 2 → 4 NO + 6 H 2 O 9. 1 B 2 Br 6 + 6 HNO 3 → 2 B(NO 3 ) 3 + 6 HBr 10. 4 NH 4 OH + 1 Kal(SO 4 ) 2 ·12H 2 O → 1 Al(OH) 3 + 2 (NH 4 ) 2 SO 4 + 1 KOH + 12 H 2 O

Tips for Balancing Equations

When balancing equations, remember chemical reactions must satisfy conservation of mass. Check your work to make certain you have the same number and type of atoms on the reactants side as on the products side. A coefficient (number in front of a chemical) is multiplied by all the atoms in that chemical. A subscript (lower number) is only multiplied by the number of atoms it immediately follows. If there is no coefficient or subscript, that is the same as a number "1" (which is not written in chemical formulas).

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Balancing Chemical Equations with Practice Problems

July 16, 2019 By Leah4sci Leave a Comment

Balancing Chemical Equations Stoichiometry Series by Leah4sci

T he videos below will show you a simple yet efficient approach to make sure you quickly account for all your numbers (minimal guessing) and don’t lose track of your work along the way.

Balancing Equations Video 1 – Introduction to Balancing with practice problems

Let’s start with the simple approach to balancing equations including practice problems so that you get comfortable with the process.

(Watch on  YouTube : Balancing Equations . Click CC for transcription.)

Balancing Combustion Reaction Equations – Balancing Video 2

Balancing combustion reactions can be tricky due to having oxygen appear in both products. Many professors solve this with a fraction but you can’t have ‘half’ oxygens in a balanced equation. Learn a simpler method for balancing combustion reactions while avoiding fractions.

(Watch on  YouTube : Balancing Combustion . Click CC for transcription.)

Balancing Acid Base Reaction Equations – Balancing Video 3

Acid Base and Ionic reactions can be balanced the same way shown above, with an additional time-saving shortcut to help you deal with all the atoms within each polyatomic ion as taught in the video below.

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<– Watch Previous Video:  Chemical Reactions –> Watch Next Video:  Coming Soon

This is Video 12 in the Stoichiometry & Reactions Video Series.  Click HERE for the entire series .

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2FeCl 3 + MgO ---> Fe 2 O 3 + 3MgCl 2 The Fe also gets balanced in this step.
2FeCl 3 + 3MgO ---> Fe 2 O 3 + 3MgCl 2 The other element (Mg or O, depending on which one you picked) also gets balanced in this step.
3Li + H 3 PO 4 ---> H 2 + Li 3 PO 4
3Li + 2H 3 PO 4 ---> 3H 2 + Li 3 PO 4 Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).
3Li + 2H 3 PO 4 ---> 3H 2 + 2Li 3 PO 4
6Li + 2H 3 PO 4 ---> 3H 2 + 2Li 3 PO 4
ZnS + (3/2)O 2 ---> ZnO + SO 2
2ZnS + 3O 2 ---> 2ZnO + 2SO 2
FeS 2 + (5/2)Cl 2 ---> FeCl 3 + S 2 Cl 2
2FeS 2 + 5Cl 2 ---> 2FeCl 3 + 2S 2 Cl 2
Fe + 3HC 2 H 3 O 2 ---> Fe(C 2 H 3 O 2 ) 3 + H 2
Fe + 3HC 2 H 3 O 2 ---> Fe(C 2 H 3 O 2 ) 3 + (3/2)H 2
2Fe + 6HC 2 H 3 O 2 ---> 2Fe(C 2 H 3 O 2 ) 3 + 3H 2
H 2 (g) + V 2 O 5 (s) ---> V 2 O 3 (s) + 2H 2 O(ℓ)
2H 2 (g) + V 2 O 5 (s) ---> V 2 O 3 (s) + 2H 2 O(ℓ)
4HCl(aq) + MnO 2 (s) ---> MnCl 2 (aq) + Cl 2 (g) + H 2 O(ℓ)
4HCl(aq) + MnO 2 (s) ---> MnCl 2 (aq) + Cl 2 (g) + 2H 2 O(ℓ)
Fe 2 O 3 (s) + C(s) ---> 2Fe(s) + CO 2 (g)
Fe 2 O 3 (s) + C(s) ---> 2Fe(s) + 3 ⁄ 2 CO 2 (g)
Fe 2 O 3 (s) + 3 ⁄ 2 C(s) ---> 2Fe(s) + 3 ⁄ 2 CO 2 (g)
2Fe 2 O 3 (s) + 3C(s) ---> 4Fe(s) + 3CO 2 (g)
2C 5 H 11 NH 2 + O 2 ---> CO 2 + 13H 2 O + NO 2
2C 5 H 11 NH 2 + O 2 ---> 10CO 2 + 13H 2 O + 2NO 2
2C 5 H 11 NH 2 + 37 ⁄ 2 O 2 ---> 10CO 2 + 13H 2 O + 2NO 2
4, 37 ---> 20, 26, 4
CO 2 + 3 ⁄ 8 S 8 ---> CS 2 + SO 2 Most of the time the fraction used to balance is something with a 2 in the denominator: 1 ⁄ 2 or 5 ⁄ 2 or 13 ⁄ 2 , for example. Not too often does one see 3 ⁄ 8 . Pretty tricky!
8CO 2 + 3S 8 ---> 8CS 2 + 8SO 2
P 4 + 3O 2 ---> 2P 2 O 3 This depends on seeing that the oxygen on the left comes in twos and the oxygen on the right comes in threes. So, you use a three and a two to arrive at six oxygens on each side. Least common multiple, baby!!
P 4 + O 2 ---> 2P 2 O 3 Then, the oxygen gets balanced: P 4 + 3O 2 ---> 2P 2 O 3
1 ⁄ 2 P 4 + O 2 ---> P 2 O 3 Let us balance the oxygen with a fraction as well: 1 ⁄ 2 P 4 + 3 ⁄ 2 O 2 ---> P 2 O 3 Finally, multiply through by two: P 4 + 3O 2 ---> 2P 2 O 3
  • Chemistry Concept Questions and Answers

Balanced Chemical Equations Questions

Chemical equations are symbolic representations of chemical reactions that express the reactants and products in terms of their respective chemical formulae. They also use symbols to represent factors such as reaction direction and the physical states of the reacting entities.

Balanced Chemical Equations Chemistry Questions with Solutions

Q1. A balanced chemical equation is in accordance with-

  • Multiple proportion
  • Reciprocal proportion
  • Conservation of mass
  • Definite proportions

Correct Answer: (c) Law of Conservation of Mass

Q2. The correct balanced equation for the reaction __C 2 H 6 O + __O 2 → __CO 2 + __H 2 O is-

  • 2C 2 H 6 O + O 2 → CO 2 + H 2 O
  • C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O
  • C 2 H 6 O + 2O 2 → 3CO 2 + 3H 2 O
  • 2C 2 H 6 O + O 2 → 2CO 2 + H 2 O

Correct Answer: (b) C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O

Q3. The correct balanced equation for the reaction __KNO 3 + __H 2 CO 3 → __K 2 CO 3 + __HNO 3 is-

  • 2KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3
  • 2KNO 3 + 2H 2 CO 3 → K 2 CO 3 + 2HNO 3
  • KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3
  • 2KNO 3 + 2H 2 CO 3 → K 2 CO 3 + 3HNO 3

Correct Answer: (a) 2KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3

Q4. The correct balanced equation for the reaction __CaCl 2 + __Na 3 PO 4 → __ Ca 3 (PO 4 ) 2 + __ NaCl is-

  • 2CaCl 2 + 2Na 3 PO 4 → 2Ca 3 (PO 4 ) 2 + NaCl
  • CaCl 2 + Na 3 PO 4 → Ca 3 (PO 4 ) 2 + NaCl
  • 3CaCl 2 + 2Na 3 PO 4 → Ca 3 (PO 4 ) 2 + 6NaCl
  • 3CaCl 2 + 2Na 3 PO 4 → Ca 3 (PO 4 ) 2 + 3NaCl

Correct Answer- (c) 3CaCl 2 + 2Na 3 PO 4 → Ca3(PO 4 ) 2 + 6NaCl

Q5. The correct balanced equation for the reaction __TiCl 4 + __H 2 O → __TiO 2 + __HCl is-

  • TiCl 4 + 2H 2 O → TiO 2 +2HCl
  • TiCl 4 + 2H 2 O → TiO 2 + 4HCl
  • 2TiCl 4 + H 2 O → 2TiO 2 + HCl
  • TiCl 4 + 4H 2 O → TiO 2 + 4HCl

Correct Answer- (b) TiCl 4 + 2H 2 O → TiO 2 + 4HCl

Q6. Write a balanced equation for the reaction of molecular nitrogen (N 2 ) and oxygen (O 2 ) to form dinitrogen pentoxide.

Answer. The equation for the reaction is-

N 2 + O 2 → N 2 O 5 (unbalanced equation)

The balanced chemical equation is-

2N 2 + 5O 2 → 2N 2 O 5

Q7. On what basis is a chemical equation balanced?

Answer. A chemical equation is balanced using the law of conservation of mass, which states that “matter cannot be created nor destroyed.”

Q8. What is the balanced equation for the reaction of photosynthesis?

Answer. The balanced chemical equation for the reaction of photosynthesis is-

6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2 .

Q9. We must solve a skeletal chemical equation.” Give a reason to justify the statement.

Answer. Skeletal chemical equations are unbalanced. Due to the law of conservation of mass, we must balance the chemical equation. It states that ‘matter cannot be created or destroyed.’ As a result, each chemical reaction must have a balanced chemical equation.

Q10. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?

Answer. The chemical equation must be balanced in order to obey the law of conservation of mass. A chemical equation is said to be balanced when the number of different atoms of elements in the reactants side equals the number of atoms in the products side. Balancing chemical equations is a trial-and-error process.

Q11. What is meant by the skeletal type chemical equation? What does it represent? Using the equation for electrolytic decomposition of water, differentiate between a skeletal chemical equation and a balanced chemical equation.

Answer. Skeletal equations are those in which formulas are used to indicate the chemicals involved in a chemical reaction.

The law of conservation of mass does not apply to skeletal equations.

The chemical formulas are represented by balanced chemical equations, which follow the law of conservation of mass, which states that the atoms on the reactant and product sides are the same.

H 2 O → H 2 + O 2 : Skeletal equation

2H 2 O → 2H 2 + O 2 : Balanced chemical equation

Q12. Write the balanced chemical equation for the following reaction:

  • Phosphorus burns in presence of chlorine to form phosphorus pentachloride.
  • Burning of natural gas.
  • The process of respiration.
  • P 4 + 10Cl 2 → 4PCl 5
  • CH 4 +2O 2 → CO 2 +2H 2 O + heat energy
  • C 6 H 12 O 6 + 6O 2 + 6H 2 O → 6CO 2 + 12H 2 O + energy

Q13. What Is the Distinction Between a Balanced Equation and a Skeleton Equation?

Answer. The primary distinction between a balanced equation and a skeleton equation is that the balanced equation provides the actual number of molecules of each reactant and product involved in the chemical reaction, whereas a skeleton equation only provides the reactants. Furthermore, a balanced equation may or may not contain stoichiometric coefficients, whereas a skeleton equation does not.

Q14. Balance the equations

  • HNO +Ca(OH) 2 → Ca(NO 3 ) 2 + H 2 O
  • NaCl + AgNO 3 → AgCl + NaNO 3
  • BaCl 2 +H 2 SO 4 → BaSO 4 +HCl

Answer. The balanced chemical equation for the reactions are as follows-

  • 2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2H 2 O
  • BaCl 2 +H 2 SO 4 → BaSO 4 + 2HCl

Q15. Write a balanced molecular equation describing each of the following chemical reactions.

  • Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
  • Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapour.
  • Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
  • Water vapour reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
  • CaCO 3 → CaO + CO 2

On heating, 1 mol of solid calcium carbonate yields 1 mol of calcium oxide and 1 mol of carbon dioxide gas.

  • 2C 4 H 10 +13O 2 → 8CO 2 + 10H 2 O When 2 moles of butane gas react with 13 moles of diatomic oxygen gas, 8 moles of carbon dioxide gas and 10 moles of water vapours are produced.
  • MgCl 2 + 2NaOH → 2NaCl + Mg(OH) 2

1 mol magnesium Cordelia reacts with two moles of sodium hydroxide to produce two moles of aqueous sodium chloride solution and one mole of solid magnesium hydroxide.

  • 2H 2 O + 2Na → 2NaOH + H 2

2 moles of water vapour react with 2 moles of sodium metal, yielding 2 moles of solid sodium hydroxide and 1 mol of hydrogen gas.

Practise Questions on Balanced Chemical Equations

Balance the following equations-

1. (NH 4 ) 2 Cr 2 O 7 (s) → Cr 2 O 3 (s) + N 2 (g) + H 2 O(g)

2. Ca(OH) 2 + H 3 PO 4 → Ca 3 (PO 4 ) 2 + H 2 O

3. FeCl 3 + NH 4 OH → Fe(OH) 3 + NH 4 Cl

4. Al 2 (CO 3 ) 3 + H 3 PO 4 → AlPO 4 + CO 2 + H 2 O

5. S 8 + F 2 → SF 6

Click the PDF to check the answers for Practice Questions. Download PDF

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How to Find Molar Ratio: Examples and Practice

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  • Last Updated On: February 12, 2024

How to find molar ratio: examples and practice problems

Chemistry isn’t just about mixing colorful liquids in beakers and waiting for an explosion—it’s a world of ratios, reactions, and relationships. At the heart of understanding this fascinating world lies the concept of the molar ratio, a key player in predicting the outcomes of chemical reactions. But what exactly is a molar ratio, and why is it so important? Simply put, it quantifies the proportions of reactants and products in a chemical equation, and accounts for every atom according to the law of conservation of mass. Remember learning about balanced equations? It’s time to put that knowledge into action. In this blog post, we’ll dive deep into the world of molar ratios, from what they are to how to find them using balanced equations. Whether you’re a budding chemist or just curious about the science behind the reactions, you’re in the right place to uncover the secrets of how to find the molar ratio.

What We Review

What is Molar Ratio?

Imagine you’re following a recipe to bake a cake. You wouldn’t just throw in random amounts of flour, sugar, and eggs, right? Just like in baking, chemistry requires precise measurements to get the desired outcome. This is where the concept of the molar ratio comes into play, acting as the “recipe” for chemical reactions.

In order to determine how much reactant must be used in an equation or how much product will be produced, molar ratio calculations are used. They are all based on the balanced equation.

A molar ratio is the proportion of moles of one substance to the moles of another substance in a chemical reaction. The coefficients of the substances in a balanced chemical equation show the molar ratio relationship. For example, in the reaction to produce water ( 2H_2 + O_2 \rightarrow 2H_2O ), the molar ratio of H_2 to O_2 is 2:1. This means that two moles of hydrogen gas react with one mole of oxygen gas to produce water.

So, why is this important? Chemists predict how much of each reactant they need to produce a certain amount of product without any waste by using molar ratio. It ensures that every atom of the reactants has a place in the products, adhering to the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

Understanding molar ratios is crucial for anyone looking to delve into the world of chemistry, whether it’s synthesizing a new compound in a research lab or figuring out the right amount of baking soda to add to your volcano science project. By mastering this concept, you’ll unlock the ability to navigate through chemical equations with ease, paving the way for exciting experiments and discoveries.

Chemical plants and other factories use balanced equations and molar ratios to determine reactant amounts to create a certain amount of product.

How to Find Molar Ratio

In order to truly grasp the concept of molar ratio, let’s dive into some example problems that highlight how to use this tool in real chemical equations. Curious about how to find the molar ratio? These examples will show you step-by-step how to calculate molar ratios and apply them to predict the amounts of reactants and products in a chemical reaction.

Example 1: Combustion of Propane

The combustion of propane ( C_3H_8 ) in oxygen ( O_2 ) is a common reaction that produces carbon dioxide ( CO_2 ) and water ( H_2O ). The balanced chemical equation for this reaction is:

Question: What is the molar ratio of O_2 to CO_2 in the combustion of propane?

Solution: To find molar ratio, look at the coefficients in your balanced equation. By looking at the balanced equation, we can see that 5 moles of O_2 produce 3 moles of CO_2 . Therefore, the molar ratio of O_2 to CO_2 is 5:3.

Example 2: Formation of Ammonia

The Haber process combines nitrogen ( N_2 ) and hydrogen ( H_2 ) gas to form ammonia ( NH_3 ), a crucial component in fertilizers. The balanced equation for this reaction is:

Question: If you start with 4 moles of N_2 how many moles of H_2 are needed to react completely based on the molar ratio?

Solution: From the balanced equation, the molar ratio of N_2 to H_2 is 1:3. This means for every mole of N_2 , they require 3 moles of H_2 to react. For 4 moles of N_2 , we need:

Therefore, 12 moles of H_2 completely react with 4 moles of N_2 .

Practice Problems: Finding and Applying Molar Ratio

Now that you’ve seen how to find molar ratio and how to use them through example problems, it’s your turn to try solving some on your own. These practice problems will test your understanding of molar ratios and how to apply them to different chemical reactions. Work through these on your own, then scroll down for solutions.

Problem 1: Synthesis of Water

When hydrogen gas ( H_2 ) reacts with oxygen gas ( O_2 ) water ( H_2O ) forms. The balanced equation for this reaction is: 

If you have 6 moles of H_2 , how many moles of O_2 are needed to react completely, and how many moles of H_2O will be produced?

Problem 2: Decomposition of Potassium Chlorate

Potassium chlorate ( KClO_3 ) decomposes upon heating to produce potassium chloride ( KCl ). The balanced equation for this reaction is:

How many moles of O_2 can be produced from the decomposition of 4 moles of KClO_3 ?

Problem 3: Combustion of Ethanol

Ethanol ( C_2H_5OH ) combusts in oxygen to produce carbon dioxide and water. The balanced chemical equation is:

If 2 moles of C_2H_5OH are combusted, how many moles of O_2 are required, and what amounts of CO_2 and H_2O are produced?

Problem 4: Production of Ammonium Nitrate

Ammonium nitrate ( NH_4NH_3 )is produced by the reaction of ammonia ( NH_4 ) with nitric acid ( HNO_3 ). The balanced chemical equation for this reaction is:

If a fertilizer company needs 5 moles of ammonium nitrate, how many moles of ammonia and nitric acid are required to achieve this?

Problem 5: Synthesis of Magnesium Oxide

Magnesium ( Mg ) reacts with oxygen ( O_2 ) to form magnesium oxide ( MgO ). The balanced equation for this reaction is:

During a lab experiment, a student reacts 6 moles of magnesium with excess oxygen. How many moles of magnesium oxide does the reaction produce, and how many moles of oxygen are consumed in the reaction?

Tips for Solving:

  • Start by identifying the molar ratios between the reactants and products from the balanced chemical equations.
  • Use the molar ratios to calculate the amounts of reactants or products as needed.
  • Remember to check your work and ensure that the law of conservation of mass is satisfied in your calculations.

Molar Ratio Practice Problem Solutions

Are you ready to see how you did? Review below to see the solutions for the molar ratio practice problems.

Remember, to find molar ratio, use the coefficients in the balanced equation. The molar ratio of H_2 to O_2 is 2:1, meaning 2 moles of H_2 react with 1 mole of O_2 . You will use this ratio to determine how many moles of O_2 are needed to react completely.

The molar ratio of H_2 to H_2O is 2:2, which can be simplified to 1:1. Therefore, if 6 moles of H_2 reacts, it produces 6 moles of H_2O .

The molar ratio of KClO_3 to O_2 is 2:3. Hence, 2 moles of KClO_3 react with 3 moles of O_2 . You will use this to determine how many moles of O_2 will be produced.

This question has three different parts. The first is determining how many moles of O_2 are needed to react completely with 2 moles of C_2H_5OH . Initially, you must find the molar ratio. The ratio of O_2 to  C_2H_5OH is 3:1.

The second part asks how much CO_2 is produced from 2 moles of C_2H_5OH . The ratio of CO_2 to C_2H_5OH is 2:1.

The third part asks how much H_2O the reaction produces from 2 moles of C_2H_5OH . The ratio of H_2O to  C_2H_5OH is 3:1.

If a fertilizer company needs to produce 5 moles of ammonium nitrate, how many moles of ammonia and nitric acid are required to achieve this?

This is a two-part problem, but solved the same way. This is because the balanced equation has a molar ratio of 1:1 for all reactants and products in the equation. Therefore, however many moles you put into the reaction, produces the same amount of moles as products. Therefore, if we want to produce 5 moles of NH_4NH_3 , we would need to put in 5 moles of NH_3 and 5 moles of HNO_3 .

During a lab experiment, a student reacts 6 moles of magnesium with excess oxygen. How many moles of magnesium oxide will be produced, and how many moles of oxygen are consumed in the reaction?

This is a two-part problem. The first part asks how many moles of Mg the reaction makes if 6 moles of Mg reacts. The molar ratio of MgO to Mg is 2:2, which can be simplified to 1:1. Therefore, if 6 moles of Mg reacts, 6 moles of MgO will be produced.

The second part of the problem asks how much excess O_2 the reaction uses if 6 moles of Mg reacted. The molar ratio of Mg to O_2 is 2:1. We use this to determine our answer.

Molar ratios aren't just something learned in chemistry class; they are used by companies across the world to make commercial products.

Embarking on the journey through the world of chemistry reveals the intricate dance of atoms and molecules, governed by fundamental principles such as the molar ratio. This concept, akin to the precise measurements in a recipe, ensures that chemical reactions proceed smoothly, with each reactant and product playing its part in the grand scheme of matter transformation.

Understanding molar ratios not only demystifies how substances react in specific proportions but also empowers us with the ability to predict the outcomes of these reactions. Whether it’s synthesizing a new compound in the lab, analyzing environmental samples, or simply marveling at the chemical reactions in everyday life, the knowledge of molar ratios serves as a crucial tool in the arsenal of any budding chemist.

In conclusion, we’ve explored how to find molar ratio and tackled practice problems to solidify our understanding. Remember, the beauty of chemistry lies not just in theoretical knowledge but in applying these concepts to solve real-world problems. So, I encourage you to continue exploring, questioning, and experimenting with the fascinating reactions that make up our world.

Chemistry continually challenges and inspires, and with tools like molar ratios, you can uncover the mysteries that lie in molecules and reactions. So, keep your curiosity alive, and let the molar ratio guide you as you journey through the incredible landscape of chemistry.

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  22. How to Find Molar Ratio: Examples and Practice

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