Reactions and moles - Higher
Limiting reactants
A reaction finishes when one of the reactantA substance that reacts together with another substance to form products during a chemical reaction. 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 limiting reactantThe reacting substance that is completely used up in a chemical reaction and which determines how much product is made.
- the reactant that is left over is described as being in excessIn chemistry, a substance is in excess if there is more than enough of it to react with another reactant.
The massThe amount of matter an object contains. Mass is measured in kilograms (kg) or grams (g). of productA substance formed in a chemical reaction. 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 stoichiometryThe ratio of the amounts of each substance in a balanced chemical equation. 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 balanced chemical equationA chemical equation written using the symbols and formulae of the reactants and products, so that the number of units of each element present is the same on both sides of the arrow.
- the mass of the limiting reactant, and
- the Ar (relative atomic massThe mean relative mass of the atoms of the different isotopes in an element. It is the number of times heavier an atom is than one-twelfth of a carbon-12 atom.) or Mr (relative formula massThe sum of the relative atomic masses of the atoms in a chemical formula.) 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 ratioA ratio is a way to compare amounts of something. It is usually written in the form a:b. 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)
| Step | Action | Result | Result |
| 1 | Write the formulae of the substances | Mg | O2 |
| 2 | Calculate the numbers of moles | \(\frac{6.08}{24.3}\) = 0.250 mol | \(\frac{4.00}{32.0}\) = 0.125 mol |
| 3 | Divide both by the smaller number of moles | \(\frac{0.250}{0.125}\) = 2 mol | \(\frac{0.125}{0.125}\) = 1 mol |
| Step | 1 |
|---|---|
| Action | Write the formulae of the substances |
| Result | Mg |
| Result | O2 |
| Step | 2 |
|---|---|
| Action | Calculate the numbers of moles |
| Result | \(\frac{6.08}{24.3}\) = 0.250 mol |
| Result | \(\frac{4.00}{32.0}\) = 0.125 mol |
| Step | 3 |
|---|---|
| Action | Divide both by the smaller number of moles |
| Result | \(\frac{0.250}{0.125}\) = 2 mol |
| Result | \(\frac{0.125}{0.125}\) = 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