Question

A sample of calcium carbonate is 80% pure. 25 gm of this sample is treated with excess of $HCl$. How much volume of $C{O}_2$ will be obtained at NTP.

• A. 4.48 litre
• B. 5.6 litre
• C. 11.2 litre
• D. 2.24 litre

Answer 1

The correct option is A 4.48 litre

$CaC{O}_3=80$%$=\frac{80}{100}\times 25=20g$

Reaction is, $\underset{1}{CaC{O}_3}\left(s\right)+2HCl\left(aq\right)⟶CaC{l}_2\left(s\right)+\underset{1}{C{O}_2}\left(g\right)+H_{2}O\left(l\right)$

Number of moles of $CaC{O}_3=\frac{20}{100}=0.2$ moles

$\therefore$ Moles of $C{O}_2=0.2$

$1$ mole occupies $22.4L$

$\therefore 0.2$ moles of $C{O}_2$ occupies= $0.2\times 22.4=4.48L$