Question

Match the Column I with Column II and choose the correct option from the codes given below.

Column I		Column II
A	$10gCaCO3\underset{\text{ Decomposition }}{\overset{\Delta }{\to }}$	1	$0.224LC{O}_2$
B	$1.06gNaN{O}_3\stackrel{\text{ Excess }HCl}{\to }$	2	$4.48LC{O}_2$
C	$2.4gC\underset{\text{ combustion }}{\overset{\text{ Excess }{O}_2}{\to }}$	3	$0.448LC{O}_2$
D	$0.56gCO\underset{\text{ combustion }}{\overset{\text{ Excess }{O}_2}{\to }}$	4	$2.24LC{O}_2$
		5	$22.4LC{O}_2$

• A. $A-(4),B-(1),C-(2),D-(3)$
• B. $A-(5),B-(1),C-(2),D-(3)$
• C. $A-(4),B-(1),C-(3),D-(2)$
• D. $A-(1),B-(4),C-(2),D-(3)$

Answer 1

Answer: (A)

The correct match is :
$A\to 4,B\to 1,C\to 2,D\to 3$
A. $\underset{100g}{CaC{O}_3}\underset{\text{ Decomposition}}{\overset{\Delta }{\to }}CaO+\underset{22.4L}{C{O}_2}$
$\because 100gCaC{O}_3$ 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. $\underset{106g}{N{a}_{2}C{O}_3}\stackrel{\text{ Excess }HCl}{\to }2NaCl+H_{2}O+\underset{22.4L}{C{O}_2}$
$\because 106gN{a}_{2}C{O}_3$ gives $=22.4LC{O}_2$
$\therefore 1.06gN{a}_{2}C{O}_3$ will give
$=\frac{22.4\times 1.06}{106}LC{O}_2$
$=0.224LC{O}_2$
C. $\underset{12}{C}\underset{\text{ Combustion }}{\overset{\text{ Excess }{O}_2}{\to }}\underset{22.4L}{C{O}_2}$
$12g$ carbon on combustion gives $=22.4LC{O}_2$
$2.4g$ carbon on combustion gives
$=\frac{22.4\times 2.4}{12}LC{O}_2$
$=2\times 2.24LC{O}_2=4.48LC{O}_2$

$56g$ carbon monoxide on combustion gives
$=2\times 22.4LC{O}_2$
$0.56g$ carbon monoxide on combustion will give
$=\frac{2\times 22.4\times 0.56}{56}=0.448LC{O}_2$