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 or Mr values of the limiting reactant and the product
Example
3.0 g of carbon reacts completely with excess oxygen to form carbon dioxide:
C(s) + O2(g) → CO2(g)
Calculate the maximum mass of carbon dioxide that can be produced. (Ar of C = 12.0, Mr of CO2 = 44.0)
\(amount\ =\ \frac{mass}{molar mass}\)
\(amount\ of\ carbon\ =\ \frac{3.0}{12.0}\)
= 0.25 mol
Looking at the equation, 1 mol of C forms 1 mol of CO2, so 0.25 mol of C forms 0.25 mol of CO2
mass = molar mass × amount
mass of CO2 = 44.0 × 0.25
= 11.0 g
Question
5.00 g of copper(II) carbonate decomposes to form copper(II) oxide and carbon dioxide:
CuCO3(g) → CuO(s) + CO2(g)
Calculate the maximum mass of carbon dioxide that can be produced. (Mr of CuCO3 = 123.5, Mr of CO2 = 44.0)
\(amount\ of\ copper(II)\ carbonate\ =\ \frac{5.00}{123.5}\)
= 0.0405 mol
Looking at the equation, 1 mol of CuCO3 forms 1 mol of CO2, so 0.0405 mol of CuCO3 forms 0.0405 mol of CO2
mass of CO2 = 44.0 × 0.0405
= 1.78 g
Stoichiometry of a reaction
The stoichiometryThe ratio of the amounts of each substance in a balanced chemical equation. 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 equation. It can be deduced or worked out using masses found by experiment.
Example
4.86 g of magnesium reacts with 3.2 g oxygen to produce magnesium oxide, MgO. Deduce the balanced equation for the reaction. (Ar of Mg = 24.3, Mr of O2 = 32.0)
| Step | Action | Result | Result |
| 1 | Write the formulae of the substances | Mg | O2 |
| 2 | Calculate the amounts | \(\frac{4.86}{24.3} = 0.20\ mol\) | \(\frac{3.2}{32.0} = 0.10\ mol\) |
| 3 | Simplify the ratio to whole numbers | 2 | 1 |
| Step | 1 |
|---|---|
| Action | Write the formulae of the substances |
| Result | Mg |
| Result | O2 |
| Step | 2 |
|---|---|
| Action | Calculate the amounts |
| Result | \(\frac{4.86}{24.3} = 0.20\ mol\) |
| Result | \(\frac{3.2}{32.0} = 0.10\ mol\) |
| Step | 3 |
|---|---|
| Action | Simplify the ratio to whole numbers |
| Result | 2 |
| Result | 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: 2Mg + O2 → 2MgO