Question

If 250 g. of CaCO3 upon reaction with excess HCl gives 222 g. of CaCl2, what is the % age purity of CaCo3?​

Answer 1

Answer:

The mass of HCl is

$\frac{0.75 \times 36.5}{1000} \times 25 = 684g$

1 mole of calcium carbonate reacts with 2 moles of HCl.

Hence, the mass of calcium carbonate that will react completely with 0.684 g of HCl is

$\frac{100}{2 \times 36.5} \times 0.684 = 0.937g.$