Chemical Formulae, Equations and Calculations Part 2
6 marks · NormalPotassium carbonate ( ) reacts with dilute hydrochloric acid:
A student adds 0.20 g of solid potassium carbonate to 25.0 cma3 of dilute hydrochloric acid (an excess is present). Calculate the number of moles of potassium carbonate used. (Ar: K = 39, C = 12, O = 16)
Calculate the maximum volume (in cma3) of carbon dioxide gas produced at room temperature and pressure. (Molar volume = 24,000 cma3 at rtp.)
After the reaction, the solution is made up to 100.0 cma3 in a volumetric flask. Calculate the concentration, in mol/dma3, of potassium chloride present in the final solution.
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) 0.00145 mol · must show number of moles calculated correctly from mass and Mr
(b) 35 cm^3 · must show moles CO2 from equation, then volume using correct molar volume
(c) 0.029 mol/dm^3 · must show calculation of mol KCl from stoichiometry, division by 0.100 dm^3
Common mistakes: Incorrect use of formula mass; Mixing up molar volumes in dm^3 and cm^3; Incorrect ratio from equation; Not converting cm^3 to dm^3 for concentration
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correct calculation of Mr of K2CO3 and setup for moles calculation (mass / Mr).
A1 (a) · 1 mark · Final answer for moles: 0.00145 mol (accept rounding to 0.0015 or 0.0014 with correct working).
M2 (b) · 1 mark · Use of molar ratio (1:1) and molar volume in setup for volume of CO2 calculation. (follow-through)
A2 (b) · 1 mark · Correct final answer for CO2 volume (35 cm^3). (follow-through)
M3 (c) · 1 mark · Correct use of stoichiometry to find moles of KCl produced and use of correct volume in dm^3. (follow-through)
A3 (c) · 1 mark · Correct final answer for concentration: 0.029 mol/dm^3. (follow-through)
Worked solution
(a) Mr of K2CO3 = (2 d 39) + 12 + (3 d 16) = 78 + 12 + 48 = 138.
Number of moles = 0.20 g / 138 g/mol = 0.00145 mol.
(b) The equation shows 1 mol K2CO3 produces 1 mol CO2. So moles of CO2 = 0.00145 mol.
Rounded to 35 cm^3.
(c) Each mole of K2CO3 produces 2 moles of KCl, so moles KCl = 2 d 0.00145 = 0.00290 mol.
The solution was made up to 100.0 cm^3 = 0.100 dm^3.