The right option for the mass of CO_2 produce | vedclass

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(C) The weight of impure limestone is $20 \ g$.Since the limestone is $20 \%$ pure,the weight of pure $CaCO_3$ is $\frac{20}{100} 	imes 20 \ g = 4 \

Vedclass · Accepted Answer

$1.76$

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(C) The weight of impure limestone is $20 \ g$.Since the limestone is $20 \%$ pure,the weight of pure $CaCO_3$ is $\frac{20}{100} 	imes 20 \ g = 4 \ g$.The molar mass of $CaCO_3$ is $40 + 12 + (3 	imes 16) = 100 \ g/mol$.The number of moles of $CaCO_3$ is $n = \frac{4 \ g}{100 \ g/mol} = 0.04 \ mol$.According to the reaction $CaCO_3 ightarrow CaO + CO_2$,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.Therefore,$0.04 \ mol$ of $CaCO_3$ produces $0.04 \ mol$ of $CO_2$.The molar mass of $CO_2$ is $12 + (2 	imes 16) = 44 \ g/mol$.The mass of $CO_2$ produced is $0.04 \ mol 	imes 44 \ g/mol = 1.76 \ g$.

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

The right option for the mass of $CO_2$ produced by heating $20 \ g$ of $20 \%$ pure limestone is $........ \ g$ (Atomic mass of $Ca = 40$) $[CaCO_3 \stackrel{1200 \ K}{\longrightarrow} CaO + CO_2]$

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