$10 g$ of a piece of marble was put into excess of dilute $H C l$ acid. When the reaction is completed $1120 c c$ of $C O_{2}$ is obtained at STP. Find % of $C a C O_{3}$ in marble.

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Solution

$1120 c m^{3} C O_{2}$
$m o l e s o f C O_{2} = m o l e s o f C a C O_{3}$
$V o l u m e o f C O_{2} i n (L) = 1120 \times \frac{1}{1000} d m^{3} (L) \Rightarrow 1.120 L$
$N o . o f m o l e s o f C O_{2} = \frac{1.120}{22.4} \Rightarrow \frac{1120}{1000 \times 22.4} \times 10$
$\Rightarrow \frac{1}{20} \Rightarrow 0.05$
$m a s s o f C a C O_{3} = 0.05 \times 100$
$\Rightarrow \frac{5}{100} \times 100 = 5 g .$
$% C a C O_{3} = \frac{5}{10} \times 100 \Rightarrow 50 %$

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