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.
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
Worked example
12 g of magnesium reacts completely with excess hydrochloric acid to form magnesium chloride and hydrogen:
Mg(s) + 2HCl(aq) 鈫 MgCl2(aq) + H2(g)
Calculate the maximum mass of hydrogen that can be produced. (Ar of Mg = 24, Mr of H2 = 2)
Amount of magnesium = \(\frac{mass}{relative~atomic~mass} \)
Amount of magnesium = \(\frac{12}{24} \)
= 0.5 mol
Looking at the equation, 1 mol of Mg forms 1 mol of H2, so 0.5 mol of Mg forms 0.5 mol of H2.
Mass of H2 = Mr 脳 amount
= 2 脳 0.5
= 1 g
Question
1.0 g of calcium carbonate decomposes to form calcium oxide and carbon dioxide:
CaCO3(g) 鈫 CaO(s) + CO2(g)
Calculate the maximum mass of carbon dioxide that can be produced. (Mr of CaCO3 = 100, Mr of CO2 = 44)
Amount of calcium carbonate = \(\frac{1.0}{100}\)
= 0.01 mol
Looking at the equation, 1 mol of CaCO3 forms 1 mol of CO2, so 0.01 mol of CaCO3 forms 0.01 mol of CO2.
Mass of CO2 = relative formula mass 脳 amount
= 44 脳 0.01
= 0.44 g
Calculating balancing numbers
The balancing numbers in an equation can be worked out using masses found by experiment.
Worked example
6.0 g of magnesium reacts with 4.0 g oxygen to produce magnesium oxide, MgO. Deduce the balanced equation for the reaction. (Ar of Mg = 24, Mr of O2 = 32)
| Step | Action | Result | Result |
| 1 | Write the formulae of the substances | Mg | O2 |
| 2 | Calculate the amounts | \(\frac{6.0}{24}\) = 0.25 mol | \(\frac{4.0}{32}\) = 0.125 mol |
| 3 | Divide both by the smaller amount | \(\frac{0.25}{0.125}\) = 2 | \(\frac{0.125}{0.125}\) = 1 |
| Step | 1 |
|---|---|
| Action | Write the formulae of the substances |
| Result | Mg |
| Result | O2 |
| Step | 2 |
|---|---|
| Action | Calculate the amounts |
| Result | \(\frac{6.0}{24}\) = 0.25 mol |
| Result | \(\frac{4.0}{32}\) = 0.125 mol |
| Step | 3 |
|---|---|
| Action | Divide both by the smaller amount |
| Result | \(\frac{0.25}{0.125}\) = 2 |
| Result | \(\frac{0.125}{0.125}\) = 1 |
This means that 2 mol of Mg reacts with 1 mol of O2, so the left-hand side of the equation is:
2Mg + O2
Then balancing in the normal way gives: 2Mg + O2 鈫 2MgO