# 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**

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

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 + O_{2} == 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*

*2SO _{2}+ O_{2} ===2SO_{3}*

*Here the mole ratio*

* 2:1:2*

*So by analysis we can boldly say*

*2 moles of SO _{2}: 1 mole of O_{2}: 2 moles of SO_{3}*

*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 + H _{2}0 ====NaOH + H_{2}*

*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 + CaCl _{2}==MgCl_{2} + 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 + O_{2} ==CO_{2}

Solution

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

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 + O_{2 }==CO_{2}

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

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

1 mole of Carbon =======1 mole of CO_{2}

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 + ZnCl*_{2}*===CaCl*_{2}* + 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 + ZnCl*_{2}*===CaCl*_{2}* + Zn* _{}

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

*Ca + ZnCl*_{2}*===CaCl*_{2}* + 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 _{4 }+ 2O_{2} === CO_{2} + 2H_{2}O*

*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 _{4 }+ 2O_{2} === CO_{2} + 2H_{2}O*

Carbon iv oxide given 1.5 mole while methane is unknown

*CH _{4 }+ 2O_{2} === CO_{2} + 2H_{2}O*

1 mole of CH_{4} =======1 mole of CO_{2}

X moles of CH_{4}====== 1.5 mole of CO_{2}

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.