Question

A sample of $Ca{CO}_3$ is 50% pure. On heating 1.12 L of ${CO}_2$ (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

Answer: (A)

7.8 g

Solution:- (A) $7.8g$

Decomposition of $CaC{O}_3$-

$CaC{O}_3⟶CaO+C{O}_2$

At STP,

$22.4L$ of a gas $=1\text{mole}$

Therefore,

$1.12L$ of $C{O}_2=\frac{1}{22.4}\times 1.12=0.05\text{mole}$

From the above reaction,

$1$ mole of $C{O}_2$ is produced by $1$ mole of $CaC{O}_3$.

$0.05$ mole of $C{O}_2$ will be produced by $0.05$ mole of $CaC{O}_3$.

Therefore,

Molar mass $CaC{O}_3=100g/mol$

Mass of $0.05$ mole of $CaC{O}_3=0.05\times 100=5g$

Now, again from the above reaction,

Molar mass of $CaO=56g/mol$

$100g$ of $CaC{O}_3$ gives $56g$ of $CaO$.

$5g$ of $CaC{O}_3$ will give $\frac{56}{100}\times 5=2.8g$

Therefore,

Total mass residue $=5+2.8=7.8g$

Hence the mass of residue left is $7.8g$.