# How is balanced chemical equation used in stoichiometry analysis?

Actually, a chemical equation is used in stoichiometry analysis by balancing the equation, comparing the mole ratio and then solving to get the unknown substance required.

Stoichiometry is defined as the quantitative study of the amount of reactants and products in a balanced chemical equation.

It is important to emphasize that stoichiometry calculations are usually hinged on balanced chemical equations.

A chemical equation is the representation of a chemical reaction using symbols and formula.

Moreover, a balanced chemical equation shows equal number of atoms on both sides of the equation which is the basis for stoichiometry calculations.

Solving stoichiometry problems require balancing chemical equation as the first step and then following the guidelines in solving the problems.

While stoichiometry focuses on. quantitative calculations involving amount of reactants and products a balanced chemical equation shows the comparisons of the substances on both sides of equitation.

## How can chemical equation be used to solve stoichiometry?

We will show how chemical equations can be used to solve stoichiometry calculations by using few examples.

Example 1

If 3.5 moles of calcium carbonate was decomposed, calculate the number of moles of calcium oxide evolved?

Solution

This question requires stoichiometry equation

Step 1

Build the stoichiometry equation

CaCO_{3}=== CaO + CO_{2}

Step 2

Compare number of moles of known and unknown substance

1 mole of CaCO_{3} == 1 mole of CaO

3.5 moles of CaCO_{3} == X

X= 3.5 X 1/ 1 = 3.5 moles of CaO

You can agree with me that this stoichiometry problem is only possible with a balanced chemical equation.

Example 2

How many moles of Nitrogen would react with hydrogen to form 4.5 moles of ammonia?

Solution

Step 1

Build the equation

N_{2} + 3H_{2} ===2NH_{3}

Step 2

Compare the number of moles

1 mole of N_{2}===2 moles of NH_{3}

X mole of N_{2}== 4 moles of NH_{3}

X= 1 mole X 4 moles /2

X = 2 moles of Nitrogen

Question 3

What mass of Zinc chloride would be produced by reacting 3 moles of Zinc with Hydrochloric acid?

Solution

Step 1

Build the equation

Zn +2HCl === ZnCl_{2}

Step 2

Compare number of moles of known and unknown

1 mole of Zinc === 1 mole of ZnCl_{2}

3 moles of Zinc== X

X= 3 mole X 1 mole/ 1 mole

X= 3 moles of Zinc Chloride

Step 3

Convert mole to mass

1 mole of ZnCl_{2} ==136 grams

3 moles if ZnCl_{2}== X

X= 3 X 186/1

X= 558 grams of Zinc chloride

Question 4

What volume of hydrogen gas is evolved by 4.6 moles of magnesium reacting with Hydrochloric

acid?

Solution

Mg + 2HCl === MgCl2 + H_{2}

1 mole of Mg === 1 mole of H_{2}

4.6 moles of Mg=== X

X= 4.6 moles X 1 mole / 1 mole

X= 4.6 moles of Hydrogen gas

Recall that

1 mole of gas at stp ==24 Litres

4.6 moles of gas at stp == X

X= 4.6 moles X 24 L/ 1 mole

X= 110.4 Litres of Hydrogen gas

Question 5

What volume of oxygen would react with Carbon to form 5.5 litres of Carbon IV oxide?

Solution

Step 1

Build the equation

C + O_{2} ===CO_{2}

Step 2

Compare the moles of Known substance and unknown substance

1 mole of carbon === 1 mole of Carbon IV oxide

Step 3

Change the moles to volume using molar volume

24 litres of O_{2} === 24 litres of CO_{2}

X litres of O_{2} === 5.5 litres of CO_{2}

_{X = 24 litres X 5.5 litres / 24 }_{litres}

_{X= 5.5 litres of Oxygen }

In conclusion , balancing chemical equation is the first step in stoichiometry analysis.