Chemical Formulae, Equations and Calculations Part 2

6 marks · Normal

Citric acid reacts with sodium bicarbonate according to the following equation:

(a)

A student reacts 0.420 g of sodium bicarbonate () with excess citric acid. Calculate the number of moles of sodium bicarbonate used. (Ar: Na = 23, H = 1, C = 12, O = 16)

[2 marks]

(b)

Calculate the volume, in dma3, of carbon dioxide gas produced at room temperature and pressure. (Molar volume = 24 dma3 at rtp.)

[2 marks]

(c)

The carbon dioxide produced is collected and completely dissolved in 250 cma3 of distilled water. Calculate the concentration of carbon dioxide (in mol/dma3) in this solution.

[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.

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(a) 0.005 mol · must show number of moles from correct Mr and mass

(b) 0.040 dm^3 · must show correct use of ratio from equation: 1 mol acid : 3 mol NaHCO3 : 3 mol CO2; then volume from moles

Common mistakes: Incorrect formula mass for NaHCO3; Wrong ratio from the balanced equation; Mixing cm^3 with dm^3 for volume

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Calculation and use of correct Mr for NaHCO3, setup for moles calculation (mass/Mr).

A1 (a) · 1 mark · Final answer for moles of NaHCO3: 0.005 mol.

M2 (b) · 1 mark · Correct use of 3:3 (or 1:1) mole ratio for CO2 and use of molar volume. (follow-through)

A2 (b) · 1 mark · Final answer for volume: 0.040 dm^3. (follow-through)

M3 (c) · 1 mark · Divide moles of CO2 by the volume in dm^3. (follow-through)

A3 (c) · 1 mark · Final answer for concentration: 0.06 mol/dm^3. (follow-through)

Worked solution

(a) Mr of = 23 + 1 + 12 + (3 \times 16) = 84.

Moles = 0.420 g / 84 g/mol = 0.005 mol.

(b) From the equation, 3 mol NaHCO3 : 3 mol CO2; so 0.005 mol NaHCO3 makes 0.005 mol CO2.

But only 1/3 of the NaHCO3 is relevant to 1 mol acid: actually, in this setup, all NaHCO3 is converted, so 0.005 mol CO2, and volume = 0.005 x 24 dm^3 = 0.12 dm^3.

Correction: every 1 mol acid gives 3 mol CO2, but since acid is in excess, all CO2 comes from 0.005 mol NaHCO3, so makes 0.005 mol CO2.

But since only 1:1 between NaHCO3 and CO2, final answer for volume is 0.005 x 24 = 0.12 dm^3.