Question

Match the following:

List-I		List-n (At STP)
(A)	$10 \, g \, CaCO_3 \xrightarrow [\text{decomposition}]{\Delta}$	(i)	$0.224 \, L \, CO_2$
(B)	$1.06 \, g \, Na_2 CO_3 \xrightarrow{\text{Excess HCl}}$	(ii)	$4.48 \, L \, CO_2$
(C)	$2.4 \, g \, C \xrightarrow [\text{combustion}]{Excess \, O_2}$	(iii)	$0.448 \, L \, CO_2$
(D)	$0.56 \, g \, CO \xrightarrow [\text{combustion}]{\text{Excess} O_2}$	(iv)	$2.24 \, L \, CO_2$
		(v)	$22.4 \, L \, CO_2$

The correct match is

• A. A $\rightarrow$ iv , B $\rightarrow$ i , C $\rightarrow$ ii , D $\rightarrow$ iii
• B. A $\rightarrow$ v , B $\rightarrow$ i , C $\rightarrow$ ii , D$\rightarrow$ iii
• C. A $\rightarrow$ iv , B $\rightarrow$ i , C $\rightarrow$ iii , D $\rightarrow$ ii
• D. A $\rightarrow$ i , B $\rightarrow$ iv , C $\rightarrow$ ii , D $\rightarrow$ iii

Answer 1

Answer: (A)

$\because 100 g CaCO _{3}$ on decomposition gives $=22.4 LCO _{2}$
$\therefore 10 g CaCO _{3}$ on decomposition will give
$=\frac{22.4 \times 10}{100} LCO _{2}$
$=2.24 L CO _{2}$
$106\, g\, Na _{2} CO _{3}$ gives $=22.4\, L\,CO _{2}$
$1.06\, g\, Na _{2} CO _{3}$ will give
$=\frac{22.4 \times 1.06}{106} L C O_{2}$
$= 0.224\, L\, C O_{2}$
$12\, g$ carbon on combustion gives $=22.4\, L\,CO _{2}$
$2.4\, g$ carbon on combustion will give
$=\frac{22.4 \times 2.4}{12} LCO _{2}$
$=2 \times 2.24 LCO _{2}$
$=4.48\, L\, CO _{2}$
$56\, g$ carbon monoxide on combustion gives
$=2 \times 22.4\, L\,CO _{2}$
$0.56\, g$ carbon monoxide on combustion will give
$=\frac{2 \times 22.4 \times 0.56}{56} LCO _{2}$
$=0.448\, L\,CO _{2}$