The molar mass of CaCO_{3} is 100 g/mol. The | vedclass

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(A) The balanced chemical equation for the thermal decomposition of $CaCO_{3}$ is:$CaCO_{3}(s) \stackrel{\Delta}{\longrightarrow} CaO(s) + CO_{2}(g)$F

Vedclass · Accepted Answer

$11$

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(A) The balanced chemical equation for the thermal decomposition of $CaCO_{3}$ is:$CaCO_{3}(s) \stackrel{\Delta}{\longrightarrow} CaO(s) + CO_{2}(g)$From the stoichiometry of the reaction,$1 \ mol$ of $CaCO_{3}$ produces $1 \ mol$ of $CO_{2}$.The molar mass of $CaCO_{3} = 100 \ g/mol$.The number of moles of $CaCO_{3}$ in $25 \ g$ is:$n(CaCO_{3}) = \frac{25 \ g}{100 \ g/mol} = 0.25 \ mol$.Since $1 \ mol$ of $CaCO_{3}$ yields $1 \ mol$ of $CO_{2}$,$0.25 \ mol$ of $CaCO_{3}$ will yield $0.25 \ mol$ of $CO_{2}$.The molar mass of $CO_{2} = 12 + (2 	imes 16) = 44 \ g/mol$.The mass of $CO_{2}$ produced is:$Mass = n 	imes M = 0.25 \ mol 	imes 44 \ g/mol = 11 \ g$.

The molar mass of $CaCO_{3}$ is $100 \ g/mol$. The maximum amount of carbon dioxide that can be liberated on heating $25 \ g$ of $CaCO_{3}$ is $.... \ g$.

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