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

$1.76$

Vedclass · Answer

(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 $0.04 \ mol 	imes 44 \ g/mol = 1.76 \ g$.

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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