Are fewer steps required in solving stoichiometry problems?

In solving stoichiometry problems, there are quite fewer steps required but there are basic things we need to understand about stoichiometry.

Though stoichiometry could be a little confusing for some students which I feel it shouldn’t anyways but I want to show you the misnomer and an easy approach to stoichiometry calculations. 

What are the steps required or taken in solving stoichiometry calculations? 

The truth is that stoichiometry calculations are of different categories 

Mole to mole stoichiometry 

Mole to mass stoichiometry 

Mole to volume stoichiometry 

Mass to mass stoichiometry 

Mass to volume stoichiometry 

Volume to volume stoichiometry 

So you need to master the art of solving these stoichiometry calculations.

5 fewer steps required in solving Stoichiometry 

Discover the type of stoichiometry problems 

The first step I recommend is for the candidate to discover what the examiner is asking.

Let me illustrate with few questions 

N2 + 3H2====2NH3

Case 1

What amount of ammonia in moles would be evolved by 4 moles of Nitrogen?

This is mole to mole stoichiometry 

Case 2 

What mass of ammonia would be evolved by 4 moles of Nitrogen ?

This is mole to mass Stoichiometry 

Case 3

What mass of ammonia would be evolved by reacting 4  grams of Nitrogen 

This is mass to mass stoichiometry 

Case 4

What volume of ammonia would be evolved by reacting 13 grams of Nitrogen. 

This is mass to volume stoichiometry 

Case 5

What volume of Nitrogen would be evolved by reacting 12cm3 of Nitrogen. 

This is volume to volume stoichiometry 

So , the first step is to deduce the type of stoichiometry problems you’re presented with. 

Check and balance the chemical equation 

Every stoichiometry calculation stems from a balanced chemical equation. 

Of Course, there is a need to learn how to balance chemical equations if you’ve not mastered the art because it is the root of stoichiometry calculations.

Introduce mole to mole ratio comparison 

When you’re certain the equation is balanced, you can thus introduce mole to mole ratio comparisons  of reactants and products.

From the mole comparisons of balanced equations , you’ve already started dismantling the stoichiometry problems. 

Please ensure you pick the coefficients of the substances because that’s the right mole to use in your calculations. 

Example 

If you are presented with this equation

2K2 + 2H2O === 2KOH + H2

Compare potassium to potassium hydroxide

2 moles of potassium ==2  moles of potassium hydroxide 

Compare potassium to hydrogen 

2 moles of potassium == 1 mole of hydrogen 

Compare water to hydrogen 

2 moles of water = 1 mole of Hydrogen

Start with the know parameter given to the unknown parameter you want to solve

At this moment, you will discover that in every stoichiometry practice questions and calculations, you need to first deduce the two substances required. All stoichiometry problems involve a given substance (reactant or product) and the unknown substance which is the one required to solve.

so from your mole ratio comparison which is the background information, you then go further to calculating the unknown or required substance.

Example 1

calculate the mass of potassium hydroxide produced when 3.3 moles of Potassium dissolves in cold water.

2K + 2H20 ====2KOH + H2

So you can see from the question that we have already gotten the amount given which is 3.3 moles of potassium , so we are looking for the mass of Potassium hydroxide that will be produced.

So  to make the calculation easier, you can underline the two substances you’re comparing .

Going forward, I will compare the two substances

background information from balanced Chemical equations

2 moles of  K ====2 moles of KOH

3.3 moles of K==== x

x = 3.3 x 2 = 6.6 moles of KOH

converting moles to mass

1 mole of KOH=56 grams of KOH

3.3 moles of KOH= x

x = 3.3 x 56 = 184.3 grams of KOH

Example 2

what mass of calcium chloride is produced by reacting 2.6 moles of Calcium according to this equation

Ca + MgCl2 =====CaCl2 + Mg

solution 

Ca + MgCl2 =====CaCl2 + Mg

background information

2 moles of  Ca====1  mole of CaCl2

2.6 moles of Ca=== x

x = 2.6/2 = 1.3  moles of CaCl2

converting moles to mass

1 mole of CaCl2= 111 grams of CaCl2

1.3 moles of CaCl2= x

x = 1.3 x 111 =144.3 grams of Calcium Chloride

Determine limiting reagent. 

The last thing I will recommend is try and understand how to determine limiting reagents. Limiting reagent or reactant is the one that finishes first.

There are some stoichiometry problems in which the amount of two reactants will be given but one will be in excess while the other one will be the limiting reactant

I have fully explained how to determine limiting reagents  before but I will just give a quick review.

Example 

Calculate the mole of zinc sulphate produced when 26 grams moles of zinc react with 250cm3 of 0.05 mol/dm3 Sulphuric acid.

Zn + H2SO4 ====ZnSO4 + H20

Solution 

compare the mole ratio of the limiting reagent and product required

You will see that there is a need to determine the limiting reagent since we are given the amount of two reactants.

Zn + H2SO4 ====ZnSO4 + H20

for zinc 

n =m/M

n= 26/65 = 0.4 mole

for H2SO4

n =cv

n= 0.05 x 250/1000

n=0.0125 mole

Then divide by the coefficients

for zinc 

0.4 /1 =0.4 

for H2SO4

0.0125/1 =0.0125 

The limiting reagent is the reactant with the least number of moles.

In conclusion, the fewer steps mentioned are helpful in solving stoichiometry problems of any degree and stoichiometry practice questions.