Calculations with equations involving solutions
2 marks · EasyA student adds 50 cm³ of 0.20 mol/dm³ hydrochloric acid, HCl, to excess calcium carbonate, CaCO₃.
The equation is:
Explain, using calculations, how to find the mass of CaCO₃ that will react with the HCl. Include your method and final answer. (Ar: Ca = 40, C = 12, O = 16)
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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0.50 g (to 2 s.f.) g (±0.01)
Also accept: 0.5 g
Common mistakes: Forgetting to convert cm³ to dm³; Using the wrong mole ratio
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Marking points
M1 · 1 mark · Calculates moles of HCl: 50 cm³ ÷ 1000 = 0.050 dm³; 0.050 × 0.20 = 0.010 mol HCl, then uses 2:1 ratio to get 0.0050 mol CaCO₃.
A1 · 1 mark · Finds mass: 0.0050 mol × 100 = 0.5 g CaCO₃ (accept 0.50 g). (follow-through)
Worked solution
Step 1: Convert 50 cm³ to dm³: 50 ÷ 1000 = 0.050 dm³.
Step 2: Calculate moles of HCl: 0.050 × 0.20 = 0.010 mol.
Step 3: From the equation, 2 mol HCl react with 1 mol CaCO₃, so 0.010 mol HCl reacts with 0.010 ÷ 2 = 0.0050 mol CaCO₃.
Step 4: Mr(CaCO₃) = 40 + 12 + (16 × 3) = 100.
Step 5: Mass = 0.0050 × 100 = 0.5 g (answer).