Calculations with equations involving solutions
5 marks · HardEthanoic acid reacts with sodium hydroxide as follows:
A student mixes 30 cm³ of 0.40 mol/dm³ ethanoic acid with excess sodium hydroxide.
Calculate the number of moles of ethanoic acid used.
Calculate the mass of sodium acetate (CH₃COONa) produced. (Relative formula mass: CH₃COONa = 82)
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) 0.012 mol
(b) 0.98 g
Common mistakes: Not converting cm³ to dm³; Not applying stoichiometric ratio
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Converts 30 cm³ to dm³ (30/1000 = 0.030 dm³).
A1 (a) · 1 mark · Correct calculation of moles: 0.030 × 0.40 = 0.012 mol.
M2 (b) · 1 mark · Uses the 1:1 mole ratio from the equation to get moles of CH₃COONa = moles of CH₃COOH. (follow-through)
M3 (b) · 1 mark · Uses mass = moles × Mr (0.012 × 82). (follow-through)
A2 (b) · 1 mark · Correct final mass for sodium acetate: 0.98 g (allow 0.984 g). (follow-through)
Worked solution
(a) Number of moles of ethanoic acid:
(b) From the equation, the ratio of CH3COOH to CH3COONa is 1:1, so moles of CH3COONa formed = 0.012 mol.
Mass = moles × Mr = 0.012 × 82 = 0.984 g
Final answer to 2 sig. fig.: 0.98 g