5 Guidelines on how to do stoichiometry?

Students’ exploration in stoichiometry has led to several approaches in stoichiometry calculations and it is very important to note the several guidelines required to do stoichiometry.

There are several guidelines on how to do stoichiometry and this will help us to solve stoichiometry calculations easily.

7 guidelines required to solve stoichiometry problems

1.Understand balancing of chemical equations

There is a need to understand how to write chemical equations and balance them.

To balance a chemical equation, we need to know that the number of atoms on both sides must be equal to each other.

Balancing of chemical equations needs simple calculations and manipulations on how to play with numbers.

Hint

You need to mention that the chemical equation is balanced because that is the first step in solving stoichiometric problems

Example of balanced chemical equations

N₂ + 3H₂ ====2NH₃

By analysis

You can see that we have 2 atoms of Nitrogen on both sides, 6 atoms of Nitrogen and 2 atoms of Hydrogen.

2Mg + O₂ ===2MgO

Also, you can see that in the above equation, we have 2 atoms of magnesium and 2 atoms of oxygen.

2. Discover how to relate moles of reactants and products:

It is important to decipher how to relate moles of reactants and products. You have to understand how to relate the number of moles of reactants and products in order to be able to solve stoichiometry questions easily.

The best way to do this comparison of moles is to underline the coefficients of the reactants and products in the balanced chemical equation.

Hint

N₂ + 3H₂ ====2NH₃

We can relate the number of moles of reactants to products

1 mole of Nitrogen ======= 2 moles of Ammonia

3 moles of hydrogen ===== 2 moles of Ammonia

3. Be able to deduce the relate moles to parameters

Then another important thing to mention is to understand the known or given parameter of the substance and the unknown or required substance in the balanced chemical equation.

Let’s analyze more with an equation

2 moles of nitrogen reacted with hydrogen. How many moles of ammonia would be produced?

    N₂ + 3H₂ ====2NH₃

Hint

The simplest step I recommend is to underline the two substances mentioned, you have to underscore the given parameter of the substance and the unknown parameter of the substance.

N₂+ 3H₂ ====2NH₃

So nitrogen and ammonia were mentioned and thus we have

1 mole of Nitrogen ======= 2 moles of ammonia

2 moles of Nitrogen ======= ?

4. Compare and solve by equivalent calculations

Since we have been able to abstract the actual number of moles of reactants and products, we can then relate and use equivalent calculations to get the answer.

Hint

N₂+ 3H₂ ====2NH₃

So nitrogen and ammonia were mentioned and thus we have

1 mole of Nitrogen ======= 2 moles of ammonia

2 moles of Nitrogen ======= x

X = 2 x 2 /1 = 4 moles of Ammonia

5. Be careful to deduce limiting reagent and excess reagent

In solving stoichiometry, sometimes you can be given the amount of two reactants and then you will be confused on the exact reactant to use and solve calculations.

When you understand how to determine limiting reagent or reactant, then it will be easier to calculate such stoichiometry problems.

Hint

If 15g of Magnesium reacted with 0.045 mol/dm³ of 200cm³  H₂SO₄. Calculate the mass of the salt (MgSO₄) produced.

2Mg + H₂SO₄ ===2MgSO₄ + H₂

I have an issue right now, I do not know the limiting reagent because I have two reactants with known data.

So you have to fully understand that you have to check the limiting reactant in the equation.

Steps in calculating limiting reagent

Step 1: First, deduce the number of moles

For magnesium

Use n = m/M

This means number of moles/ amount = reacting mass/Molar mass

So n= 15/24 =0.625 mole

For Sulphuric acid

Use n=cv

This means number of moles = concentration in mol/dm³ x volume in dm³

So n= 0.045 x 200/1000= 0.009 mole

Step 2: To finally deduce the limiting reagent, divide by the number of moles of the reactants in the balanced equation

Number of moles /coefficients

For magnesium

0.0625 /2 = 0.03125

For sulphuric acid

0.009/1 = 0.009

So, the limiting reagent is the reactant with the least or smallest number of moles and this  sulphuric acid. Then we can now solve this equation

Hint

You should solve with the actual number of moles and not the number gotten by dividing by the coefficients.

So finally, we will solve like thus

2Mg + H₂SO₄ ===2MgSO₄ + H₂

1 mole of sulphuric acid =======2 moles of magnesium sulphate

0.009 mole of sulphuric acid ===== x

X= 0.009 x 2 /1 = 0.018 mole of MgSO₄

These guidelines on stoichiometry will help you figure out how to solve simple problems involving stoichiometry.