Excess reactants

A good way to ensure that one reactant fully reacts is to use an excess of the other reactant. This is financially efficient when one of the reactants is very cheap.

When one reactant is in excess, there will always be some left over. The other reactant becomes a limiting factor and controls how much of each product is produced.

While using excess reactants can help to increase percentage yields, this is at the expense of atom economy.

A balance between the economic and environmental value of the use of excess reactants must be established.

Example one

What volume of CO₂ would be produced when 5 g of calcium carbonate (CaCO₃) reacts with excess hydrochloric acid (HCl)? (molar volume = 22.4 litres mol^-1)

Start with a balanced equation and look at the molar ratio. We know the acid is in excess, so the number of moles of calcium carbonate that react will control how many moles of product are formed.

\(2HCl + CaCO_{3}\,\,\,\rightarrow \,\,\,\,CaCl_{2} + H_{2}O + CO_{2}\)

\(\,\,\,\,\,\,\,\,\,\,\,\,1\,mole\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,mole\)

The number of moles of CaCO3 equals the mass divided by the formula mass. This is 5 divided by 100, which equals 0.05 moles.

\(1\,\,mole\,\,CaCO_{3}\rightarrow 1\,\,mole\,\,CO_{2}\)

\(0.05\,\,moles\,\,CaCO_{3}\rightarrow 0.05\,\,moles\,\,CO_{2}\)

\(1\,\,mole\,\,CO_{2}=22.4\,\,litres\)

\(0.05\,\,moles\,\,CO_{2}=\frac{0.05}{1}\times 22.4\)

\(=1.12\,\,litres\)

Example two

Which reactant is in excess when 0.25 g of magnesium is added to 100 cm³ of 0.1 mol l^-1 hydrochloric acid (HCl)?

Start with a balanced equation.

\(Mg + 2HCl \rightarrow MgCl_{2} + H_{2}\)

The equation shows that one mole of magnesium will react with two moles of acid. Use the information in the question to calculate how many moles of each reactant are present.