Question

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

Column I		Column II
A	$10\, g\, CaCO 3 \xrightarrow[\text { Decomposition }]{\Delta}$	1	$0.224\, L\, CO _{2}$
B	$1.06\, g\, Na NO _{3} \xrightarrow{\text { Excess } HCl }$	2	$4.48\, L\, CO _{2}$
C	$2.4\, g C \xrightarrow[\text { combustion }]{\text { Excess }O_{2}}$	3	$0.448\, L\, CO _{2}$
D	$0.56\, g\, CO \xrightarrow[\text { combustion }]{\text { Excess } O _{2}}$	4	$2.24\, L\, CO _{2}$
		5	$22.4\, L\, CO _{2}$

• 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 :
$A \rightarrow 4, B \rightarrow 1, C \rightarrow 2, D \rightarrow 3$
A. $\underset{100\, g }{CaCO _{3}} \xrightarrow[\text { Decomposition}]{\Delta} CaO +\underset{22.4\, L }{ CO _{2}}$
$\because 100\, g\, CaCO _{3}$ on decomposition gives $=22.4\, L\, CO _{2}$
$\therefore 10\, g\, CaCO _{3}$ on decomposition will give
$=\frac{22.4 \times 10}{100} L\, CO _{2}=2.24\, L\, CO _{2}$
B. $\underset{106\,g}{Na _{2} CO _{3}} \xrightarrow{\text { Excess } HCl }2 NaCl + H _{2} O + \underset{22.4\,L}{CO _{2}}$
$\because 106\, g\, Na _{2} CO _{3}$ gives $=22.4\, L\, CO _{2}$
$\therefore 1.06\, g\, Na _{2} CO _{3}$ will give
$=\frac{22.4 \times 1.06}{106} L\, CO _{2}$
$=0.224\, L\, CO _{2}$
C. $\underset{12}{ C } \xrightarrow[ \text { Combustion }] {\text { Excess } O _{2}}\underset{22.4\, L } {CO _{2}}$
$12\, g$ carbon on combustion gives $=22.4\, L\, CO _{2}$
$2.4\, g$ carbon on combustion gives
$=\frac{22.4 \times 2.4}{12} L\, CO _{2}$
$=2 \times 2.24\, L\, CO _{2}=4.48\, L\, CO _{2}$

$56\, g$ carbon monoxide on combustion gives
$=2 \times 22.4\, L\, CO _{2}$
$0.56\, g$ carbon monoxide on combustion will give
$=\frac{2 \times 22.4 \times 0.56}{56}=0.448\, L\, CO _{2}$