Balanced Equations and Reacting Masses

Chemical equations are the foundation of quantitative chemistry. This lesson covers balancing equations, using mole ratios to calculate reacting masses, identifying limiting reagents, and calculating percentage yield and atom economy.

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Balancing Chemical Equations

A balanced equation has the same number of atoms of each element on both sides. This follows the law of conservation of mass — matter cannot be created or destroyed in a chemical reaction.

Steps for Balancing

1. Write the correct formulae for all reactants and products.
2. Balance one element at a time, starting with the most complex species.
3. Leave hydrogen and oxygen until last.
4. Check that all elements are balanced.
5. Ensure state symbols are included: (s), (l), (g), (aq).

Examples of Balanced Equations

Combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Reaction of sodium with water: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Thermal decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g)

Neutralisation of sulfuric acid by sodium hydroxide: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

Exam Tip: State symbols are essential in many exam mark schemes. Always include them. Remember: (aq) means dissolved in water — aqueous solution.

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Mole Ratios and Reacting Mass Calculations

The coefficients in a balanced equation give the mole ratio of reactants and products.

Method for Reacting Mass Calculations

1. Write the balanced equation.
2. Calculate the moles of the substance you know (using n = m/M).
3. Use the mole ratio to find the moles of the substance you want.
4. Convert back to mass (using m = n × M) or volume as required.

Worked Example 1: Simple Reacting Mass

Calculate the mass of carbon dioxide produced when 5.00 g of calcium carbonate is decomposed.

CaCO₃(s) → CaO(s) + CO₂(g)

Step 1: n(CaCO₃) = 5.00 / 100.1 = 0.04995 mol

Step 2: From the equation, 1 mol CaCO₃ produces 1 mol CO₂. n(CO₂) = 0.04995 mol

Step 3: m(CO₂) = 0.04995 × 44.0 = 2.20 g

Worked Example 2: Different Mole Ratios

Calculate the mass of water produced when 4.60 g of ethanol (C₂H₅OH) undergoes complete combustion.

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

n(C₂H₅OH) = 4.60 / 46.0 = 0.100 mol

From the equation: 1 mol C₂H₅OH produces 3 mol H₂O.

n(H₂O) = 0.100 × 3 = 0.300 mol

m(H₂O) = 0.300 × 18.0 = 5.40 g

Worked Example 3: Finding Volume of Gas Produced

Calculate the volume of hydrogen gas (at RTP) produced when 0.460 g of sodium reacts with excess water.

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

n(Na) = 0.460 / 23.0 = 0.0200 mol

From the equation: 2 mol Na produces 1 mol H₂.

n(H₂) = 0.0200 / 2 = 0.0100 mol

V(H₂) = 0.0100 × 24.0 = 0.240 dm³ (or 240 cm³)

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Limiting Reagent

In most reactions, one reagent is used up first — this is the limiting reagent. The other reagent(s) are said to be in excess. The limiting reagent determines the maximum amount of product that can be formed.

Method for Identifying the Limiting Reagent

1. Calculate the moles of each reactant.
2. Divide each by its coefficient in the balanced equation.
3. The reactant with the smallest value is the limiting reagent.

Worked Example 4: Identifying the Limiting Reagent

5.40 g of aluminium reacts with 16.0 g of iron(III) oxide. Which is the limiting reagent?

2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

n(Al) = 5.40 / 27.0 = 0.200 mol → 0.200 / 2 = 0.100 n(Fe₂O₃) = 16.0 / 159.7 = 0.1001 mol → 0.1001 / 1 = 0.1001

Al gives a smaller value (0.100), so aluminium is the limiting reagent.

The maximum moles of Fe produced = 0.200 mol (same ratio as Al, 2:2).

m(Fe) = 0.200 × 55.8 = 11.2 g

Worked Example 5: Limiting Reagent with Excess