Question

The amount of 50% pure H2SO4 required to react completely with 50g of CaCO3 is?
a) 49g
b) 98g
c) 24g
d) 38g

Answer 1

Answer: The correct answer is Option b.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  ....(1)

Given mass of calcium carbonate = 50 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in above equation, we get:

$\text{Moles of }CaCO_3=\frac{50g}{100g/mol}=0.5mol$

For the reaction of sulfuric acid and calcium carbonate, the equation follows:

$CaCO_3+H_2SO_4\rightarrow CaSO_4+H_2O+CO_2$

By Stoichiometry of the reaction:

1 mole of calcium carbonate reacts with 1 mole of sulfuric acid.

So, 0.5 moles of calcium carbonate will react with = $\frac{1}{1}\times 0.5=0.5mol$ of sulfuric acid

Mass of sulfuric acid reacted will be found out by using equation 1:

$0.5=\frac{\text{Mass of sulfuric acid}}{98g/mol}\\\\\text{Mass of sulfuric acid}=49g/mol$

But, in the question, it is asked that only 50% of sulfuric acid is given, so mass of sulfuric acid will be required more by the amount of = $\frac{49\times 100}{50}=98g$

Hence, the correct answer is Option b.