Mass Calculations in Balanced Equations and Limiting Reactants

Having learned how to convert between relative masses and moles, and to balance chemical equations, we’re now equipped to calculate various aspects of chemical reactions.

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• One useful application is predicting the masses of products from the masses of reactants.

Limiting and Excess Reactants

It’s important to remember one important factor when determining the actual masses or moles of substances in a reaction:

It’s rare for an experiment to start with the precise amounts of reactants needed for every molecule to react completely. More often, one of the reactants will be used up first. This reactant is known as the limiting reactant.

In any chemical reaction, the limiting reactant is the one that determines the maximum amount of product that can be formed. Once the limiting reactant is depleted, the reaction cannot proceed because there is no more of this reactant to react with the other substances, so the reaction ends.

The excess reactant is the one that is not fully consumed in the reaction. It’s left over because there was more of it than was necessary to fully react with the limiting reactant.

When calculating the mass of the products formed in a reaction, knowing the mass of the limiting reactant is often enough. This is because the mass of the product depends directly on the amount of the limiting reactant. Therefore, even if other reactants are present in excess, they do not affect the total mass of the product formed.

Let’s look at two different examples:

Mg (s) + O₂ (g) → MgO (s)

If we burn 5g of Mg in an oxygen-rich atmosphere, how much MgO will be produced?

1. Identify the limiting reactant.

Given that there is an excess of oxygen, magnesium (Mg) is the limiting reactant.

2. Balance the equation.

Mg (s) + O₂ (g) → MgO (s)

Mg (s) + O₂ (g) → 2MgO (s)

So the balanced equation is:

2Mg (s) + O₂ (g) → 2MgO (s)

3. Convert the quantity of magnesium from masses to moles.

Molar masses of compounds are calculated by summing the relative atomic masses of the elements in the compound from the periodic table.

• Remember to take into account the number of atoms of each element in the compound!

0.206 mol

4. Identify the number of moles of product generated.

From the balanced equation, 2Mg (s) + O2 (g) → 2MgO (s), we observe a 1:1 mole ratio between Mg and MgO.

Therefore, the reaction will produce 0.206 moles of MgO.

5. Convert the number of moles of the product to mass.

• Molar mass of MgO = 24.31 (Mg) + 16 (O) = 40.31 g/mol

• Mass of MgO produced = 0.206 moles × 40.31 g/mol = 8.30g

Therefore, the reaction yields 8.30g of magnesium oxide.

This example involves a non-enclosed system. The measured mass may appear to increase as the reaction takes place because the added mass comes from the oxygen in the air.

It won’t always be clear which reactant is the limiting one. However, with the masses of each reactant known, you can determine the limiting reactant.

We can identify the limiting reactant by calculating the moles of each reactant involved in the reaction equation. Let’s look at the steps to solve this.

If 2.8g of Al are reacted with 4.25g of Cl2, find the limiting reactant using the equation:

Al (s) + Cl₂ (g) → AlCl₃ (s)

1. Balance the equation:

Al (s) + Cl₂ (g) → AlCl₃ (s)

Al (s) + 3Cl₂ (g) → 2AlCl₃ (s)

Therefore, the balanced equation is: 2Al (s) + 3Cl₂ (g) → 2AlCl₃ (s)

2. Convert quantities to moles:

Use the molar masses calculated from the periodic table:

For Aluminium (Al):

0.104 mol

For Chlorine (Cl₂):

0.060 mol

3. Identify the limiting reactant:

From the balanced equation, a 2:3 ratio of Al to Cl2 is needed to produce AlCl3. Therefore:

• For every 2 moles of Al, 3 moles of Cl₂ are required.

• With 0.104 mol of Al available, we would need 0.156 mol of Cl₂ to fully react with all the Al (based on the 2:3 ratio), but only 0.060 mol of Cl₂ is available.

Therefore, Cl₂ is the limiting reactant since it will be consumed first, limiting the amount of AlCl₃ produced.