What will be the volume of CO_2 at NTP obtain | vedclass

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(B) The chemical reaction for the heating of limestone is: $CaCO_3(s) 	o CaO(s) + CO_2(g)$Given mass of limestone = $10 \ g$.Since it is $90\%$ pure,

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$2.016$

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(B) The chemical reaction for the heating of limestone is: $CaCO_3(s) 	o CaO(s) + CO_2(g)$Given mass of limestone = $10 \ g$.Since it is $90\%$ pure,the mass of pure $CaCO_3 = 10 	imes 0.90 = 9 \ g$.Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g/mol$.Number of moles of $CaCO_3 = \frac{9 \ g}{100 \ g/mol} = 0.09 \ mol$.According to the stoichiometry of the reaction,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$.Therefore,$0.09 \ mol$ of $CaCO_3$ will produce $0.09 \ mol$ of $CO_2$.At $NTP$,the volume of $1 \ mol$ of any gas is $22.4 \ L$.Volume of $CO_2 = 0.09 \ mol 	imes 22.4 \ L/mol = 2.016 \ L$.

What will be the volume of $CO_2$ at $NTP$ obtained on heating $10 \ g$ of ($90\%$ pure) limestone (in $L$)?

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