Question

Calcium carbonate decomposes on heating according to the equation;
$CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$
At STP, the volume of $C{O}_2$ obtained by thermal decomposition of 50 g of $CaC{O}_3$ will be:

• A. 22.4 litre
• B. 44 litre
• C. 11.2 litre
• D. 1 litre

Answer 1

The correct option is C 11.2 litre

$C_{a}C{O}_3\left(s\right)\to C_{a}O\left(s\right)+C{O}_2\left(g\right)$

$1mol1mol$

Molecular weight of $CaC{O}_3$ is 100g.

$So,atSTP,$ $100g{C}_{a}C{O}_{3}gives22.4litreofC{O}_2$.

$So,50g{C}_{a}C{O}_{3}gives=\frac{22.4\times 50}{100}$

$=11.2litre$

Hence, option $C$ is correct.