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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 $\frac{22.4 	imes 10}{1

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

Vedclass · Answer

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

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

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