IMAT 2014 Q49 [Chemical Reaction]

The relative molecular mass of calcium carbonate is 100.

What is the minimum volume of 2.0 M hydrochloric acid that would be needed to completely react with 2.0 g of calcium carbonate?

A. 40 cm^3
B. 20 cm^3
C. 30 cm^3
D. 5 cm^3
E. 10 cm^3

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Let’s start by writing down the reaction:

CaCO_3 + 2HCl → CaCl + CO_2 + H_2O

From this, we can see our molar ratio in terms of reactants. 2 moles of HCl reacts with every 1 mole of Calcium Carbonate.

If 1 mole of Calcium Carbonate is 100g, x is 2.0g. X is 0.02 moles of calcium carbonate.

We can put this into a new ratio. If 2 moles of HCl reacts with 1 mole of CaCO_3, x moles of HCl reacts with 0.02 moles of CaCO_3.

X is 0.04 moles of hydrochloric acid.

If c=nV,

2M= 0.04V
V= 0.04/2
V= 0.02 dm^3 or Litre

1 L= 1000 ml
1 ml= 1 cm^3

0.02 L=20 ml
20 ml= 20 cm^3

And our answer is B.