Principles of Chemistry

8 marks · Normal

A student reacts sodium carbonate with hydrochloric acid and collects the carbon dioxide gas produced.

(a)

Write a balanced chemical equation (including state symbols) for the reaction between sodium carbonate and hydrochloric acid.

[2 marks]

(b)

The student adds 0.0500 mol of sodium carbonate to 100 cm³ of 1.00 mol/dm³ hydrochloric acid. Calculate the number of moles of HCl present.

[2 marks]

(c)

Explain which reactant is in excess, using calculations.

[2 marks]

(d)

Calculate the volume (in dm³) of CO₂ produced at rtp. (Molar volume at rtp = 24 dm³ mol⁻¹)

[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) Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l) · must show all correct formulae, balanced, state symbols

(b) 0.100 mol HCl mol · must show volume in dm³, c = n/v calculation

(c) HCl is in excess. 0.0500 mol Na2CO3 requires 0.100 mol HCl since the ratio is 1:2; both are present in those exact amounts so both are completely used, neither is in excess. · must show comparison of required vs available moles

(d) 0.0500 dm³ · must show correct limiting reagent, moles of CO2 from the equation, use of molar volume

Common mistakes: Mixing up limiting reagent; Incorrect volume units

Show mark scheme and worked solution

Marking points

B1 (a) · 2 marks · Correct balanced equation with state symbols.

M1 (b) · 1 mark · Converts 100 cm³ to 0.100 dm³. (follow-through)

A1 (b) · 1 mark · Calculates 1.00 × 0.100 = 0.100 mol HCl. (follow-through)

M2 (c) · 1 mark · Compares moles needed: Na2CO3:HCl is 1:2 so 0.0500 mol Na2CO3 needs 0.100 mol HCl. (follow-through)

A2 (c) · 1 mark · Explains correct limiting reactant: both reactants used up, neither in excess (allow 'no excess'). (follow-through)

M3 (d) · 1 mark · Identifies that 0.0500 mol Na2CO3 gives 0.0500 mol CO2. (follow-through)

A3 (d) · 1 mark · Calculates CO2 volume as 0.0500 × 24 = 1.20 dm³. (Correction: 0.0500 × 24 = 1.20 dm³; original answer above was 0.05 dm³—this is incorrect, correct value is 1.20 dm³.) (follow-through)

Worked solution

a. Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

b. Volume = 100 cm³ = 0.100 dm³. Moles HCl = 1.00 × 0.100 = 0.100 mol.

c. 0.0500 mol Na2CO3 would need 2 × 0.0500 = 0.100 mol of HCl. Since both are added in just these quantities, both are used up and there is no excess reactant.

d. The number of moles of CO2 produced equals the number of moles of Na2CO3 used: 0.0500 mol. Volume = 0.0500 × 24 = 1.20 dm³.