Question

How much amount of $CaC{O}_3$ in grams having percentage purity $50$ percent produces $0.56$ $\text{L}$ of $C{O}_2$ at STP on heating?

• A. $2.5\text{g}$
• B. $7\text{g}$
• C. $5\text{g}$
• D. $3.5\text{g}$

Answer 1

The correct option is C $5\text{g}$

We know one mole of any substance at STP has a volume of $22.4$ $\text{L}$. So $0.56\text{L of}C{O}_2\text{means}\frac{1}{40}\text{moles of}C{O}_2$
$CaC{O}_3\to CaO+C{O}_2$
By stoichiometry
$1$ mole $CaC{O}_3\to$$1$ mole $C{O}_2$
$\frac{1}{40}\text{mole of}CaC{O}_3\to \frac{1}{40}\text{moles of}C{O}_2$
Molar mass of $CaC{O}_3=100\text{g}$
So,$\text{mass of pure}CaC{O}_3\text{required}=\frac{1}{40}\times 100=2.5\text{g}$
But here since purity of $CaC{O}_3$ is $50$ percent, amount of $CaC{O}_3$ required $=2.5\text{g}\times 2=5\text{g}$