# How is mole ratio used in Stoichiometry?

Mole ratio in Stoichiometry is basically the guiding principle used in calculating Stoichiometry problems.

It is important to recall that mole concept is the Stoichiometry principle used in calculating Stoichiometry questions.

Moreover, in calculating stoichiometry problems you need to first relate the use the mole ratio and then be able to compare the substances and then relate it to either mass, volume or number of particles.

## How is mole ratio used in Stoichiometry Equation?

Mole ratio is simply the ratio of the moles of substances in a balanced chemical equation.

Example 1

N2 +3H2 ===2NH3

Here the mole ratio for Nitrogen, Hydrogen and Ammonia is

1:3:2

The ratio is gotten from the coefficients of the reactants and products

And we can use this mole ratio to solve some simple problems

By analysis we can relate

1:     3:    2

10g: 30g: 20g

Also

We can relate to other parameters like volume and number of particles using mole

Example 2

Consider the combustion of magnesium with oxygen.

2Mg + O2 == 2MgO

The mole ratio here is

2: 1: 2

Comparatively

20g : 10g : 20g

Example 3

Consider the production of sulphur vi oxide. Determine the mole ratio

2SO2+ O2 ===2SO3

Here the mole ratio

2:1:2

So by analysis we can boldly say

2 moles of SO2: 1 mole of O2: 2 moles of SO3

Mole ratio and stoichiometry calculations

We then need to learn how to apply mole ratio to solving stoichiometry problems using some stoichiometry practice questions.

Example 4

What number of moles of sodium hydroxide alkali is produced by the dissolution of 3.5 moles of sodium in water?

Solution

Na + H20 ====NaOH + H2

Solution

1 mole of Na == 1 mole of NaOH

3.5 moles of Na === X

X= 3.5 X 1/1

×= 3.5 moles of sodium hydroxide

Example 5

Study the reaction equations

Mg + CaCl2==MgCl2 + Ca

What’s the comparison here?

1:1:1:1

## Mole ratio and solving Stoichiometry Problems

Actually, mole ratio comparison is applied in solving stoichiometry problems.

With mole ratio, one can easily solve a lot if stoichiometry calculations.

This is because you can easily relate mole ratio to volume, mass and number of particles.

Okay, let’s study with some examples

Mole ratio and mass calculations

Example 1

What is the mass of carbon iv oxide evolved in this reaction if 3.4 moles of carbon reacting with oxygen?

C + O2 ==CO2

Solution

C + O2 ==CO2

So I’ll start by getting the mole ratio of the two substances required fir the calculation.

Hint: In every stoichiometry calculations, you need to know the two substances required for the calculation.

The simply way to do this is to know the substance given and the substance required to calculate.

So from this equation   C + O2 ==CO2

Carbon was given as 3.5 moles while carbon iv oxide is required

C + O2 =======CO2

1 mole of Carbon =======1 mole of CO2

3.5 moles of Carbon ====== X

X = 3.5 x 1/1 = 3.5 moles of carbon iv oxide

Example 2

What is the mole of Zinc produced by reacting 0.9 mole of Calcium?

Ca + ZnCl2===CaCl2 + Zn

Again the way to go around this is to know the substance given and the substance required to calculate.

So from this equation   Ca + ZnCl2===CaCl2 + Zn

Calcium was given as 0.9 mole while mole of zinc is required

Ca + ZnCl2===CaCl2 + Zn

1 mole of Calcium =======1 mole of Zn

0.9 mole of Calcium====== X

X = 1 x 1/1 = 1 mole of zinc

Example 3

What’s the mole of methane that will produce 1.5 moles of carbon iv oxide according to this equation?

CH+ 2O2 === CO2 + 2H2O

Solution

Again the way to go around this is to know the substance given and the substance required to calculate.

So from this equation

CH+ 2O2 === CO2 + 2H2O

Carbon iv oxide given 1.5 mole while methane is unknown

CH+ 2O2 === CO2 + 2H2O

1 mole of CH4 =======1 mole of CO2

X moles of CH4====== 1.5 mole of CO2

X = 1 x 1.5 /1= 1.5 moles of Methane

Then we can convert further to mass

I mole of methane = 16 grams

1.5 moles of methane = x

X= 1.5 x 16

X = 24 grams of methane

In conclusion, mole ratio in stoichiometry is very important and is widely employed in solving different degrees of stoichiometry problems.