# How do I calculate Mole to Mole stoichiometry?

In mole-to-mole stoichiometry, there is a need to understand how to compare the relationship between two or more substances in an equation.

This enables us to calculate the amount of mole of substances required or produced

It is important to understand mole to mole relationships and calculations but I will recommend starting from mole to mole comparisons

Stoichiometry calculations is based simply on *mole concept* and the understanding of this concept enables us to do our calculations easily..

## Examples of Mole-to-Mole Calculations

This is one of the basic calculations you can get in the stoichiometry table.

**Example 1**

Calculate the number of moles of sulphur vi oxide evolved by reacting 3.5 moles sulphur iv oxide and oxygen according to this equation

2SO_{2} + O_{2}====2SO_{3}

Solution

*First, compare the two substances mentioned in the equation*

*Secondly, compare the moles of the two substances mentioned.*

*Thirdly, underline the moles of the two substances mentioned*

*In this case, it is SO2 and SO3*

2SO_{2} + O_{2}====2SO_{3}

2 moles of SO_{2} = 2 moles of SO_{3}

3.5 moles of SO_{2} = X

X = 3.5 X 2 /2 = 3.5 moles of SO_{3}

Therefore, 3.5 moles of sulphur vi oxide evolved

**Example 2**

Consider the equation;

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

If 3.3 moles of Nitrogen reacted with unknown mole of hydrogen. Calculate the number of moles of ammonia evolved.

Solution

*Again, first you compare the substances involved*

*Secondly, compare the moles of the two substances mentioned.*

*Thirdly, underline the moles of the two substances mentioned*

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

So we have to calculate based on number of moles

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

3.3 moles of N_{2}= X

X= 3.3 X 2 /1

X= 6.6 moles of ammonia

Therefore 6.6 moles of ammonia evolved when 3.3 moles of nitrogen were evolved

**Example 3**

How many moles of Propane is needed to yield 5 moles of carbon iv oxide according to this

C_{3}H_{8} + 5O_{2}===3CO_{2} + 4H_{2}O

Solution

C_{3}H_{8} + 5O_{2}===3CO_{2} + 4H_{2}O

1 mole of C_{3}H_{8} ==3 moles of CO_{2}

X moles of C_{3}H_{8}==5 moles of CO_{2}

X= 1 mole X 5 moles /3 moles

× = 1.667 mole of carbon iv oxide evolved

**Example 4**

What is the number of moles of carbon iv oxide produced when 0.65 moles of oxygen complexity burns an unknown mass of carbon.

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

I will go straight to the calculations since I have laid down the principle a second time.

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

0.65 mole of O_{2}===X

X= 0.65 X 1 = 0.65 mole of CO_{2}

So therefore 0.65 mole of carbon iv oxide evolved

**Example 5**

If 3.35 moles calcium carbonate decomposed completely to yield calcium oxide and carbon iv oxide, what’s the number of moles of carbon iv oxide produced?

Solution

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

If 1 mole of CaCO_{3}==1 mole of CO_{2}

3.35 moles of CaCO3 ==×

X= 3.35 x1 = 3.35 moles of CO_{2}

So 3.35 moles of carbon iv oxide evolved

In conclusion, mole to mole stoichiometry is a basic calculation in stoichiometry based on mole concept analysis.