Stoichiometry and Limiting Reagents

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Calculating Stoichiometry

You need to be able to calculate the stoichiometry of an equation given the masses of reactants and products

Suppose we are given the equation….

Calcium carbonate → Calcium oxide + Carbon dioxide

…. And we know that 100g of calcium carbonate decomposed fully to give 56g of calcium oxide.

According to the principal of conservation of mass, mass cannot be created or destroyed, only converted from one form to another.

Therefore, the 44g that appear to be ‘lost’ in the decomposition of calcium carbonate must have formed carbon dioxide for mass to have been conserved.

Now consider the relative formula masses of each of the compounds involved in the reaction:

Calcium carbonate: 100

Calcium oxide: 56

Carbon dioxide: 44

Clearly, the reaction has produced one mole each of calcium oxide and carbon dioxide from one mole of calcium carbonate. So the ratio of these compounds must be 1:1:1.

Therefore, we now know that the balanced equation must look like:

\(1CaCO_3 \rightarrow 1CaO + 1CO_2 \)

Or simply:

\(CaCO_3 \rightarrow CaO + CO_2 \)

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Calculating Stoichiometry

Now consider the reaction…..

Nickel (III) oxide → Nickel + Oxygen

… And suppose that 330.8g of nickel(III) oxide decomposes to give 234.8g of nickel.

According to the principle of conservation of mass, \(330.8 – 234.8 = 96g\) of oxygen must have been produced.

Relative formula masses of compound:

Nickel (III) oxide: 165.4

Nickel: 58.7

Oxygen: 32 (Remember, oxygen exists naturally as a homoatomic molecule, \(O_2\), so its relative formula mass is \( 2\times 16=32\))

Therefore, two moles of nickel oxide yields four moles of nickel and three moles of oxygen.

So the balanced symbol equation is:

\(2Ni_2O_3 \rightarrow 4Ni + 3O_3\)

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Limiting and Excess Reagents

Limiting reagent: The substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.

Excess reactant: The reactant in a chemical reaction with greater amount than necessary to react completely with the limiting reagent.

Example 1: Consider the combustion of methane in the laboratory. The limiting reagent is methane, the excess reactant is oxygen - the combustion is not going to use up all the supply of oxygen in the room.

Example 2: Consider the equation \[4Al +3O_2 \rightarrow 2Al_2O_3\]

Formula weights:

Aluminium: 27

Oxygen: 32

Imagine you have 64g of oxygen and 54g of aluminium. Which is the limiting reagent?

64g of oxygen is 2 moles. 2 moles of oxygen would require 8/3 moles or 72g of aluminium to fully react.

54g of aluminium is also 2 moles. 2 moles of aluminium would require 3/2 moles or 24g of oxygen to react.

Therefore, there is enough oxygen for all of the aluminium to be reacted with, but not enough aluminium for all of the oxygen to be fully reacted with. Therefore, the limiting reagent must be the aluminium.