Question

Match the following :

	List-I		List-II (At STP)
(A)	$10\,g \,CaCO_3\xrightarrow[\text{decomposition}]{\Delta} $	(i)	$0.224\, L \,CO _{2}$
(B)	$1.06\,g \,NaCO_3\xrightarrow{\text{Excess} HCl}$	(ii)	$4.48\, L\, CO _{2}$
(C)	$2.4\,g \,C\xrightarrow [\text{combustion}]{\text{Excess}\, O _{2}}$	(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 - iv , B - i, C - ii, D - iii
• B. A - v , B - i , C - ii , D - iii
• C. A - iv , B - i , C - iii , D - ii
• D. A - i , B - iv , C - ii , D - iii

Answer 1

Answer: (C)

(A)$Ca\underset{100\,g}{CO_3}\xrightarrow [\text{decomposition}]{\Delta}CaO + \underset{22.4\,L}{CO_2}$
$\because \,100 \,g \,CaCO _{3}$ on decomposition gives
$=22.4\, L\, CO _{2}$
$\therefore \,10\, g \,CaCO _{3}$ on decomposition will give
$=\frac{22.4 \times 10}{100} LCO _{2} $
$=2.24\, L \,CO _{2}$
(B) $\underset{106\,g}{Na _{2} CO _{3}} \xrightarrow{\text { Excess HCl }}2 NaCl + H _{2} O +\underset{22.4 \, L }{ CO _{2}}$
$106 \, g \,Na _{2} CO _{3}$ gives $=22.4 \, L \,CO _{2}$
$1.06 \,g \, Na _{2} CO _{3}$ will give
$=\frac{22.4 \times 1.06}{106} \, L \, CO _{2} $
$=0.224 \, L \,CO _{2}$
(C) $\underset{12\, g }{ C }\xrightarrow[\text{combustion}]{\text{Excess} O _{2}} \underset{22.4\,L}{CO_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}\, L\,CO _{2} $
$=2 \times 2.24\, L \,CO _{2}$
$=4.48 \,L\,CO _{2}$
(D)$\underset{\underset{56\,g}{2[12+16]}}{2CO}\xrightarrow[\text{combustion}]{\text{Excess} O _{2}} \underset{2\times22.4\,L}{2CO_2}$
$56 \,g$ carbon monoxide on combustion
gives $=2 \times 22.4 \,L \,CO _{2}$
$0.56 \,g$ carbon monoxide on combustion will give
$=\frac{2 \times 22.4 \times 0.56}{56}\, L\,CO _{2}$
$=0.448\, L\,CO _{2}$
Hence, $A -( iv ), B -( i ), C -( ii ), D -( iii )$