Reactions and moles - Higher

Limiting reactants

A reaction finishes when one of the is all used up. The other reactant has nothing left to react with, so some of it is left over:

• the reactant that is all used up is called the
• the reactant that is left over is described as being in

The of formed in a reaction depends upon the mass of the limiting reactant. This is because no more product can form when the limiting reactant is all used up.

Balanced chemical equations

A balance chemical equation shows the ratio in which substances react and are produced. This is called the of a reaction.

For example:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

This balanced chemical equation shows that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to form 1 mole of magnesium chloride and 1 mole of hydrogen.

Reacting mass calculations

The maximum mass of product formed in a reaction can be calculated using:

• the
• the mass of the limiting reactant, and
• the Ar () or Mr () values of the limiting reactant and the product

Calculating the mass of products

Example

2.43 g of magnesium reacts completely with excess hydrochloric acid to form magnesium chloride and hydrogen:

Mg(s) + 2HCl(aq) $$\rightarrow$$ MgCl2(aq) + H2(g)

Calculate the maximum mass of hydrogen that can be produced. (Relative masses: Mg = 24.3, H2 = 2.0)

number of moles of Mg = $$\frac{2.43}{24.3}$$

= 0.100 mol

Looking at the balanced chemical equation, 1 mol of Mg forms 1 mol of H2, so 0.100 mol of Mg forms 0.100 mol of H2

mass of H2 = Mr × number of moles of H2

= 2.0 × 0.100

= 0.20 g (to 2 significant figures)

Example

1.0 g of calcium carbonate decomposes to form calcium oxide and carbon dioxide:

CaCO3(g) $$\rightarrow$$ CaO(s) + CO2(g)

Calculate the maximum mass of carbon dioxide that can be produced. (Relative formula masses: CaCO3 = 100.1, CO2 = 44.0)

number of moles of CaCO3 = $$\frac{mass}{M_r}$$

= $$\frac{1.0}{100.1}$$

= 0.010 mol (to 2 significant figures)

Looking at the balanced chemical equation, 1 mol of CaCO3 forms 1 mol of CO2, so 0.010 mol of CaCO3 forms 0.010 mol of CO2

mass of CO2 = Mr x number of moles of CO2

= 44.0 × 0.010

= 0.44 g

Calculating the mass of reactants

Example

Magnesium reacts with oxygen to produce magnesium oxide:

2Mg(s) + O2(g) $$\rightarrow$$2MgO(s)

Calculate the mass of magnesium needed to produce 12.1g of magnesium oxide.

(Relative masses: Mg = 24.3, MgO = 40.3)

number of moles of MgO = $$\frac{mass}{M_r}$$

= $$\frac{12.1}{40.3}$$

= 0.300 mol

Looking at the balanced chemical equation, 2 mol of MgO forms from 2 mol of Mg, so 0.300 mol of MgO forms from 0.300 mol of Mg

mass of Mg = Mr × number of moles of Mg

= 24.3 × 0.300

= 7.29 g

Stoichiometry of a reaction

The stoichiometry of a reaction is the of the amounts of each substance in the balanced chemical equation. It can be deduced or worked out using masses found by experiment.

Example

6.08 g of magnesium reacts with 4.00 g oxygen to produce magnesium oxide, MgO.

Deduce the balanced equation for the reaction. (Relative masses: Mg = 24.3, O2 = 32.0)

StepActionResultResult
1Write the formulae of the substancesMgO2
2Calculate the numbers of moles$\frac{6.08}{24.3}$ = 0.250 mol$\frac{4.00}{32.0}$ = 0.125 mol
3Divide both by the smaller number of moles$\frac{0.250}{0.125}$ = 2 mol$\frac{0.125}{0.125}$ = 1 mol

This means that 2 mol of Mg reacts with 2 mol of O2, so the left hand side of the equation is:

2Mg + O2

Then balancing in the normal way: 2Mg + O2 → 2MgO

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