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(A) $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. $100 \ g \ CaCO_3$ gives $22.4 \ L \ CO_2$ at $STP$. So, $10 \ g \ CaCO_3$ gives $2.24 \ L \ CO_2$. Thus

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$A-(iv), B-(i), C-(ii), D-(iii)$

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(A) $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. $100 \ g \ CaCO_3$ gives $22.4 \ L \ CO_2$ at $STP$. So, $10 \ g \ CaCO_3$ gives $2.24 \ L \ CO_2$. Thus, $A-(iv)$.$(B)$ $Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$. $106 \ g \ Na_2CO_3$ gives $22.4 \ L \ CO_2$. So, $1.06 \ g \ Na_2CO_3$ gives $0.224 \ L \ CO_2$. Thus, $B-(i)$.$(C)$ $C + O_2 ightarrow CO_2$. $12 \ g \ C$ gives $22.4 \ L \ CO_2$. So, $2.4 \ g \ C$ gives $(22.4 	imes 2.4) / 12 = 4.48 \ L \ CO_2$. Thus, $C-(ii)$.$(D)$ $2CO + O_2 ightarrow 2CO_2$. $56 \ g \ CO$ gives $2 	imes 22.4 \ L \ CO_2 = 44.8 \ L \ CO_2$. So, $0.56 \ g \ CO$ gives $(44.8 	imes 0.56) / 56 = 0.448 \ L \ CO_2$. Thus, $D-(iii)$.Therefore, the correct match is $A-(iv), B-(i), C-(ii), D-(iii)$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

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