# How Do I Solve Gas Stoichiometry Problems?

Stoichiometry gas problems involves different categories and also involve gas stoichiometry, but the approach used same.

There is a step-by-step way of solving stoichiometry problems which is still the same for stoichiometry involving gases.

## What’s gas stoichiometry?

*Gas stoichiometry is the simply the particular stoichiometry concept that deals with gases in a balanced chemical equation.*

In solving gas stoichiometry, it is still valid to recall that stoichiometry can be done very easily especially if you follow the rules.

**The simple rules include;**

*Write a balanced chemical equation**Compare the mole ratios of reactants and products in the balanced chemical equation**Use the known substance to determine the unknown substance*

I’m going to work you through some examples of gas stoichiometry problems.

Yes, I’m aware of some gas stoichiometry problems involving Ideal gas law since temperature and pressure are at different conditions.

I’m going to show all the possible stoichiometry problems.

## Gas Stoichiometry Problems

Actually, the gas stoichiometry solved problems is just to lay emphasis on the simple steps involved in solving stoichiometry.

**Task 1**

N2 + 3H2====2NH3

If 3.4 moles of Nitrogen react with excess Hydrogen.

Calculate the mole and volume of ammonia evolved.

**Solution**

N2+ 3H2===2NH3

*Check if the chemical equation is balanced*

*Compare the mole ratio of Nitrogen and ammonia*

1 mole of N2===2 moles of NH3

3.4 moles of N2===X

X= 3.4 X 2 /1 = 6.8 moles of NH3 evolved

Volume evolved

*You can use stoichiometry conversion here*

If 1 mole of NH3===22.4dm3

6.8 moles of NH3==X

X= 6.8 X 22.4 =152.32 dm3 of ammonia evolved

**Task 2**

Calculate the mole and mass of carbon iv oxide evolved from the reaction of 2 3 moles of Hydrochloric acid and calcium carbonate .

*Check if the chemical equation is balanced*

*Compare the mole ratio of Hydrochloric acid and calcium carbonate*

CaCO3 + 2HCl====CaCl2 + H2O + CO2

**Solution**

CaCO3 + 2HCl===CaCl2+ H2O +CO2

2 moles of HCl===1 mole of CO2

2.3 moles of HCl=== X

X= 2.3 /2= 1.15 mole of CO2

So, you use stoichiometry conversion to change to mass

I mole of CO2 == 44 grams

1.15 moles of CO2 == X

X= 1.15 X 44 =50.6 grams of carbon iv oxide

**Task 3**

If 4.5 moles of zinc react with Hydrochloric acid.

Calculate the mole and volume of hydrogen evolved according to this equation.

Zn + 2HCl====ZnCl2 + H2

**solution**

1 mole of Zn ==== 1 mole of H2

4.5 moles of Zn===X

X== 4.5 X 1 /1

X= 4.5 moles of Hydrogen

Since 1 mole of H2==22.4 dm3

4.5 moles of H2==X

X= 4.5 X 22.4 = 100.8dm3 of Hydrogen gas

**Task 4**

If 15.8g of magnesium reacts with Zinc Chloride

Calculate the mole , mass and volume of chlorine evolved .

Mg + ZnCl2 ===MgCl2 + Cl2

**Solution**

1 mole of Mg ==1 mole of Cl2

Change mass to mole

I mole of Mg==24 grams

X ==15.8 grams

X== 1 mole X 15.8 / 24

X==0.658 mole of Mg

Comparing the moles

1 mole of Mg ==1 mole of Cl2

0.658 mole of Mg== X

X= 0.658 X 1 = 0.658 mole of Chlorine

Converting to volume

1 mole of Cl2=== 22 4 dm3

0.658 mole of Cl2== X

X== 0.658 X 22.4

X=14.74 dm3 if Chlorine

Converting to mass

1 mole of Cl2== 71 grams

0.658 mole of Cl2== X

X= 0.658 X 71

X=46.72 grams of Chlorine gas

**Task 5**

If 3.3 moles of Nitrogen burnt in oxygen at stp according to this equation

N2 + O2===2NO

Calculate the volume of Nitrogen ii oxide at 660mmHg and 13.6 C.

**Solution**

Using the equation of reaction

At stp

1 mole of N2 ===2 moles of NO

3.3 moles of N2 ==X

X= 3.3 X 2 /1 = 6.6 moles of Nitrogen II oxide

Convert to volume

1 mole of NO== 22.4 dm3

6.6 moles of NO=X

X= 6.6 x22.4 =147.84 dm3

Using general gas equation

P1V1/T1= P2V2/T2

V2= P1V1T2/P2T1

Using the data

P1 =760 mmHg

V1=147.84 dm3

T1= 273K

P2=660 mmHg

T2=286.6K

V2=?

Substituting the values

V2= P1V1T2/P2T1

V2= 760x 147.84x 286.6/600× 273

V2=760× 147.84×286.6/600×273

V2=196.6 dm3

So, in general gas stoichiometry is very crucial in general stoichiometry but the only important thing is that it involves quantities of gaseous reactants and products.