Principles of Chemistry

8 marks · Normal

This question is about writing balanced equations (with state symbols) and using them to perform gas volume calculations.

(a)

Magnesium reacts with steam to produce magnesium oxide and hydrogen gas. Write a balanced chemical equation for this reaction, including state symbols.

[2 marks]

(b)

Calculate the maximum volume, in cm³, of hydrogen gas formed at rtp when 3.60 g of magnesium is reacted with excess steam. (Ar: Mg = 24.0; molar volume at rtp = 24,000 cm³ mol⁻¹)

[4 marks]

(c)

State one reason why the actual volume of hydrogen collected in an experiment might be less than your calculated answer.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) Mg(s) + H2O(g) → MgO(s) + H2(g) · must show correct reactants and products, balanced equation, all correct state symbols

(b) 3.60 g Mg produces 2240 cm³ H2 at rtp cm³ · must show calculation of moles of Mg, correct mole ratio used, calculation of volume of H2, correct units

(c) Some hydrogen may escape before collection or not all magnesium reacts completely. · must show loss of gas, incomplete reaction, measurement error

Show mark scheme and worked solution

Marking points

B1 (a) · 1 mark · Correct balanced equation (formulae and stoichiometry).

B2 (a) · 1 mark · All correct state symbols.

M1 (b) · 1 mark · Calculation of moles of Mg. (follow-through)

M2 (b) · 1 mark · Correct use of stoichiometry to find moles of H2. (follow-through)

M3 (b) · 1 mark · Correct calculation of H2 volume using molar volume (24,000 cm³ mol⁻¹). (follow-through)

A1 (b) · 1 mark · Final answer for H2 volume with correct units and significant figures (allow error carried forward). (follow-through)

B3 (c) · 1 mark · Correct relevant reason for loss/lower yield (e.g. gas loss, incomplete reaction).

B4 (c) · 1 mark · Alternative relevant reason (e.g. side reactions, leaks, measurement error).

Worked solution

a. Mg(s) + H2O(g) → MgO(s) + H2(g)

b. Step 1: Moles of Mg = mass / Ar = 3.60 g / 24.0 = 0.150 mol

Step 2: From the balanced equation, 1 mol Mg produces 1 mol H2. So, 0.150 mol Mg produces 0.150 mol H2.

Step 3: Volume of H2 = moles × molar volume = 0.150 × 24,000 = 3,600 cm³.

Correction: 3.60 g / 24.0 = 0.150, (0.150 mol × 24,000 = 3,600 cm³), not 2240 cm³. Thus, the maximum volume is 3,600 cm³.

c. The actual volume can be lower because some gas may escape before being collected or not all magnesium reacts.