# How Can a Chemical Equation be used to Solve Stoichiometry Problems?

A chemical equation is always used in solving stoichiometry problems because stoichiometry itself involves the relationships between reactants and products in a balanced chemical equation.

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

Since stoichiometry involves the calculations involving comparison between reactants and products, a balanced chemical equation is necessary in solving stoichiometric problems.

### What are chemical equations and reaction stoichiometry?

Chemical equations are simply symbolic representations of chemical reactions in form of symbols and chemical formulae while reaction stoichiometry is the calculation involving the relationship between reactants and products.

There is a strong relationship between balanced chemical equations and reaction stoichiometry. As a result, stoichiometry problems and its solution are solely based on the comparison of reactants and products in a balanced chemical equation.

### How do you solve stoichiometry step by step?

Recall that a balanced chemical equation shows the stoichiometric coefficients, which are the relative amounts of each substance in the reaction which is the same as number of moles of reactants and products.

To solve stoichiometry problems using a balanced chemical equation, the following steps should be followed:

- Crosscheck the balanced chemical equation for the reaction.

- Then deduce the known and unknown quantities. (The known quantities are the quantity given in the problem, while the unknown quantity is what is required)

- Compare the stoichiometric coefficients(moles) from the balanced equation to determine the mole ratio of the substances in the balanced equation and then relate them to the known substance and the unknown substance.

- Calculate the moles of the unknown substance based on the mole ratio

- Convert the moles of the unknown substance to the desired unit, for example if mass, number of particles or volume is required then, convert the given number of moles to the desired quantities using the molar mass, Avogadro’s number or molar volume respectively

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

A balanced chemical equation can be used in solving stoichiometry problems simply by comparing the reactants and products of the balanced chemical reaction and thus solving the stoichiometry calculations following the step-by-step rule of stoichiometry.

### How do you solve stoichiometry Problems?

Solving stoichiometry Problems simply requires understanding of the concept guiding the calculations and closely following the rules and solving the calculations.

Solving stoichiometry Problems could be made very easy whether it is mass-mass stoichiometry, mole-mole stoichiometry, mass-volume stoichiometry etc and the proper understanding is vital in Chemistry since stoichiometry cuts across various facets of chemistry.

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

Examples.

### Stoichiometry Problems and Answers

Example 1

What mass of Magnesium chloride could be produced by reacting 3.3 moles of Magnesium with Hydrochloric acid.

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

**Solution **

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

*Step 1: Crosscheck the equation *

The equation is crosschecked, and it is balanced

*Step 2: Deduce and compare the moles of two substances mentioned in the stoichiometry Problem*

Magnesium chloride and Magnesium are mentioned in the question, so underline the coefficients

_Mg + 2HCl =====_MgCl_{2 }+ H_{2}

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

3.3 moles of Mg ==== X

*Step 3: Use the known substance to get the unknown substance*

X = 3.3 moles X 1 mole / 1 mole

X= 3.3 moles of MgCl_{2}

*Step 4: Convert the mole to mass*

Finally, there is a need to get the mass of the magnesium chloride, so convert the mole to mass

We will the molar mass to convert

1 mole of MgCl_{2} ==== 95 grams

3.3 moles of MgCl_{2} === X

X = 3.3 ~~moles~~ x 95 grams /1 ~~mole~~ = 313.5 grams of MgCl_{2}

Example 2

If 2.4 moles of sodium react vigorously in cold water to form an alkali and liberate hydrogen.

What volume of hydrogen gas would be evolved?

2Na + 2H_{2}O====2NaOH + H_{2}

**Solution**

*Step 1: Crosscheck the equation *

The equation is crosschecked, and it is balanced

*Step 2: Deduce and compare the moles of two substances mentioned in the stoichiometry Problem*

Sodium and hydrogen mentioned in the equation, so underline the coefficients

2Na + 2H_{2}O====2NaOH + _H_{2}

2 moles of Na ==== 1 mole of H_{2}

2.4 moles of Na ==== X

*Step 3: Use the known substance to get the unknown substance*

X = 2.4 moles X 1 mole / 2 moles

X= 1.2 moles of H_{2}

*Step 4: Convert the mole to volume*

Finally, there is a need to get the volume of hydrogen gas evolved, so convert the mole to volume

We will the molar volume to convert

1 mole of H_{2} ==== 22.4 dm^{3}

1.2 moles of H_{2} === X

X = 1.2 ~~moles~~ x 22.4 dm^{3 }/1 ~~mole~~ = 26.88 dm^{3} or 26.88 litres of Hydrogen gas

Example 3

What mass of Ammonia is evolved by the reaction of 12 grams of nitrogen with hydrogen?

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

**Solution**

*Step 1: Crosscheck the equation *

The equation is crosschecked, and it is balanced

*Step 2: Deduce and compare the moles of two substances mentioned in the stoichiometry Problem*

Nitrogen and Ammonia mentioned in the equation, so underline the coefficients.

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

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

*Step 3: Use the known substance to get the unknown substance*

We are given the mass of Nitrogen to get the mass of Ammonia.

So we compare the moles of the two substances but since mass is given, change the moles to mass using molar mass.

*Step 4: Compare the masses of the reactants and products from the balanced chemical equation*

*In this case, we are given mass so convert the moles to mass*

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

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

28 grams of N_{2 }=== (2 x 17) grams of NH_{3}

28 grams of N_{2} ==== 34 grams of NH_{3}

12 grams of N_{2} ==== X

X= 14.57 grams of H_{2}

You can see that the foundation of all our Stoichiometry calculations started from the balanced chemical equation because the stoichiometry calculation starts from the comparison between the substances in the balanced chemical equation.

Additionally, the stoichiometry problems I used to illustrate are from different categories of stoichiometry Problems. I made a post on 50 stoichiometry problems to help you improve your skills and even a step-by-step stoichiometry worksheet for practice questions.