Question

$100 \,g$ impure $CaCO _{3}$ on heating gives $5.6 \,L . CO _{2}$ gas at NTP. Find the percentage of calcium in the limestone sample. [At. wt.: $Ca =40, C =12, O =16$ ]

Answer 1

$CaCO _{3} \longrightarrow CaO + CO _{2}$
$\frac{5.6}{22.4}=\frac{1}{4} mol$
Mol of $CaO = mol$ of $Ca =\frac{1}{4}$
Mass of $Ca =\frac{1}{4}, \times 40=10$
$\%$ of $Ca$ in sample $=\frac{10}{100} \times 100=10 \%$