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(B) The chemical reaction for the decomposition of limestone is: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$Calculate the mass of pure $CaCO_3$ in $150 \

Vedclass · Accepted Answer

$63$

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(B) The chemical reaction for the decomposition of limestone is: $CaCO_3(s) ightarrow CaO(s) + CO_2(g)$Calculate the mass of pure $CaCO_3$ in $150 \ kg$ of limestone:$	ext{Mass of } CaCO_3 = 150 \ kg 	imes 0.75 = 112.5 \ kg = 112500 \ g$Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g \ mol^{-1}$Number of moles of $CaCO_3 = \frac{112500 \ g}{100 \ g \ mol^{-1}} = 1125 \ mol$According to the stoichiometry of the reaction,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CaO$.Therefore,moles of $CaO = 1125 \ mol$Molar mass of $CaO = 40 + 16 = 56 \ g \ mol^{-1}$Mass of $CaO = 1125 \ mol 	imes 56 \ g \ mol^{-1} = 63000 \ g = 63 \ kg$

The amount of calcium oxide produced on heating $150 \ kg$ of limestone ($75 \%$ pure) is $........... \ kg$ $(Nearest \ integer)$. Given: Molar mass (in $g \ mol^{-1}$) of $Ca = 40, O = 16, C = 12$.

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