Question

Match the following :

	List-I		List-II (At STP)
(A)	$10gCaC{O}_3\underset{\text{decomposition}}{\overset{\Delta }{\to }}$	(i)	$0.224LC{O}_2$
(B)	$1.06gNaC{O}_3\stackrel{\text{Excess}HCl}{\to }$	(ii)	$4.48LC{O}_2$
(C)	$2.4gC\underset{\text{combustion}}{\overset{\text{Excess}{O}_2}{\to }}$	(iii)	$0.448LC{O}_2$
(D)	$0.56gCO\underset{\text{combustion}}{\overset{\text{Excess}{O}_2}{\to }}$	(iv)	$2.24LC{O}_2$
		(v)	$22.4LC{O}_2$

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)$Ca\underset{100g}{C{O}_3}\underset{\text{decomposition}}{\overset{\Delta }{\to }}CaO+\underset{22.4L}{C{O}_2}$
$\because 100gCaC{O}_3$ on decomposition gives
$=22.4LC{O}_2$
$\therefore 10gCaC{O}_3$ on decomposition will give
$=\frac{22.4\times 10}{100}LC{O}_2$
$=2.24LC{O}_2$
(B) $\underset{106g}{N{a}_{2}C{O}_3}\stackrel{\text{ Excess HCl }}{\to }2NaCl+H_{2}O+\underset{22.4L}{C{O}_2}$
$106gN{a}_{2}C{O}_3$ gives $=22.4LC{O}_2$
$1.06gN{a}_{2}C{O}_3$ will give
$=\frac{22.4\times 1.06}{106}LC{O}_2$
$=0.224LC{O}_2$
(C) $\underset{12g}{C}\underset{\text{combustion}}{\overset{\text{Excess}{O}_2}{\to }}\underset{22.4L}{C{O}_2}$
$12g$ carbon on combustion gives $=22.4LC{O}_2$
$2.4g$ carbon on combustion will give
$=\frac{22.4\times 2.4}{12}LC{O}_2$
$=2\times 2.24LC{O}_2$
$=4.48LC{O}_2$
(D)$\underset{\underset{56g}{2[12+16]}}{2CO}\underset{\text{combustion}}{\overset{\text{Excess}{O}_2}{\to }}\underset{2\times 22.4L}{2C{O}_2}$
$56g$ carbon monoxide on combustion
gives $=2\times 22.4LC{O}_2$
$0.56g$ carbon monoxide on combustion will give
$=\frac{2\times 22.4\times 0.56}{56}LC{O}_2$
$=0.448LC{O}_2$
Hence, $A-(iv),B-(i),C-(ii),D-(iii)$