Question

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

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

Answer 1

Answer: (C)

11.2 litre

No.of moles of $CaC{O}_3$ = $\frac{50}{100}$ $=0.5$

According to reaction, 1 mole of $CaC{O}_3$ gives 1 mole of $C{O}_2$.

So 0.5 moles of $CaC{O}_3$ will give 0.5 moles of $C{O}_2$

Now at STP 1 mole of any gas = 22.4 L

So 0.5 moles of $C{O}_2$ gas $=$ 22.4$\times$0.5 = 11.2 L