Question

A sample of $CaC{O}_3$ is $50\%$ pure. On heating $1.12L$ of $C{O}_2$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is:

• A. 7.8 g
• B. 3.8 g
• C. 2.8 g
• D. 8.9 g

Answer 1

The correct option is A 7.8 g

No. of moles of $C{O}_2$ evolved $=\frac{1.12}{22.4}=0.05$ $moles$

$CaC{O}_3\left(s\right)\stackrel{\Delta }{⟶}CaO\downarrow +C{O}_2\uparrow$

$0.05$ $0.05$ $moles$

So, $0.05$ $moles$ of $CaC{O}_3$ have reacted.

Mass $=0.05\times 100=5$ $gm=50\%$ of $CaC{O}_3$ sample

Total weight $=2\times 5$ $gm=10$ $gm$

Residue left by $CaC{O}_3=5$ $gm$

Residue left by $CaO=56\times 0.05=2.8$ $gm$

Toatl residue $=5+2.8=7.8$ $gm$