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Question 22 of 48 in Carboxylic Acids

Reaction with metals

2 marks · Easy

A student reacts 0.81 g of magnesium with excess butanoic acid. Calculate the maximum volume of hydrogen gas produced at room temperature and pressure. (1 mol of any gas occupies 24 dm³ at r.t.p.; Mg = 24.3 g/mol)

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0.40 dm³ (±0.02)

Common mistakes: Dividing by 12.15 or using wrong molar mass; Missing multiplication by 24

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Marking points

M1 · 1 mark · Finds moles of magnesium: moles = mass / Mr = 0.81 / 24.3 = 0.0333 mol. (follow-through)

A1 · 1 mark · Multiplies by 24 to obtain final answer: 0.0333 mol × 24 dm³ = 0.80 dm³ for 2 moles H₂, but considering the correct 1:1 ratio from the balanced equation, gives the correct volume of 0.40 dm³.

Worked solution

Step 1: Calculate moles of magnesium:

Step 2: From the equation: 1 mole of Mg gives 1 mole of H₂ gas.

Step 3: Volume of hydrogen = 0.0333 mol × 24 dm³ = 0.8 dm³ (for 1 mole, not 2).

However, double-checking, actually of Mg produces 0.0333 mol of H₂. Volume = 0.0333 × 24 = 0.80 dm³

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