Reversing the calculations
5 marks · HardCalcium hydroxide solution reacts with hydrochloric acid according to the following balanced equation:
In a titration, 25.0 cm³ of hydrochloric acid solution of concentration 0.120 mol/dm³ is reacted completely with calcium hydroxide solution.
Calculate the number of moles of HCl present in 25.0 cm³ of solution.
Determine the number of moles of Ca(OH)₂ required to react with this amount of HCl.
Given that the concentration of Ca(OH)₂ is 0.0800 mol/dm³, calculate the volume (in cm³) of Ca(OH)₂ solution that supplies this amount of Ca(OH)₂.
Explain briefly why it is important to use the balanced equation when carrying out these calculations.
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) 0.00300 mol
(b) 0.00150 mol
(c) 18.8 cm³
(d) Because the balanced equation gives the correct stoichiometric ratio between reactants, which shows how many moles of each reactant react together. · must show States importance of correct stoichiometric ratio
Common mistakes: Not converting cm³ to dm³; Using wrong ratio in stoichiometry; Incorrect formula rearrangement; Ignoring mole ratios
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Convert 25.0 cm³ to dm³ and multiply by HCl concentration to find moles.
M2 (b) · 1 mark · Use the 2:1 ratio from the balanced equation to find moles of Ca(OH)₂. (follow-through)
M3 (c) · 1 mark · Rearrange concentration formula to solve for volume in dm³. (follow-through)
A1 (c) · 1 mark · Correctly convert to cm³ and give answer to 3 significant figures. (follow-through)
B1 (d) · 1 mark · States that the balanced equation shows the correct ratio in which reactants combine.
Worked solution
(a) Moles of HCl:
(b) According to the equation, 2 mol HCl reacts with 1 mol Ca(OH)₂, so required moles of Ca(OH)₂:
(c) Volume of Ca(OH)₂ required (in dm³):
Convert to cm³: (rounded to 3 significant figures)
(d) The balanced equation gives the correct ratio between reactants so you can calculate the right amount of each needed for the reaction.