Question

10 gram of a piece of marble was put into excess of dilute HCl acid. When the reaction was complete, $1120c{m}^3$ of $O_2$ was obtained S.T.P. the percentage of $CaC{O}_3$ in the marble is:

• A. 10%
• B. 25%
• C. 50%
• D. 75%

Answer 1

The correct option is C 50%

$CaC{O}_3+2HCl\to CaC{l}_2+H_{2}O+C{O}_2$

100g or one mole 24000ml

100 g of $CaC{O}_3$ gives 22400 ml of carbon dioxide.

10 g of $CaC{O}_3$ gives = $\frac{22400}{100}\times 10=2240$ ml of carbon dioxide.

If the metal is 100% pure, 2240 ml of carbon dioxide is produced.

1120 ml of carbon dioxide is produced in this case.

So the % purity $=\frac{100}{2240}\times 1120=50$%

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