Question

Chalk is almost pure calcium carbonate. We can work out its purity by measuring how much carbon dioxide is given off when it reacts with $HCl$. $15g$ of chalk was reacted with an excess of dilute hydrochloric acid. 2.24 liters of carbon dioxide gas was collected at standard temperature and pressure (STP).
The equation for the reaction is :
$CaC{O}_3\left(s\right)+2HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$
Calculate the percentage purity.
(Assume that impurity present does not produce $C{O}_2$)

• A. $40.77\%$
• B. $57.52\%$
• C. $66.67\%$
• D. $78.33\%$

Answer 1

The correct option is C $66.67\%$

Reaction involved :-
$CaC{O}_3\left(s\right)+2HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)$
From the balanced chemical reaction, 1 mole of $CaC{O}_3$ produces 1 mole of $C{O}_2$

We know, 1 mole of gas has a volume of 22.4 liters at STP.

Thus, $22.4L$ of $C{O}_2$ gas is produced by $100g$ of $CaC{O}_3$
By unitary method $2.24L$ of gas is produced by $\frac{100}{22.4}\times 2.24$ = 10 g of $CaC{O}_3$
$Percentagepurity=\frac{Massofpuresample}{Massofimpuresample}\times 100$
= $\frac{10}{15}\times 100=66.67\%$