150 g CaCO_3 sample was taken. On its comple | vedclass

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(B) The decomposition reaction is: $CaCO_3 \rightarrow CaO + CO_2$ Molar mass of $CaCO_3 = 100 \ g/mol$ Molar mass of $CaO = 56 \ g/mol$ From the stoi

Vedclass · Accepted Answer

$66.67$

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(B) The decomposition reaction is: $CaCO_3 ightarrow CaO + CO_2$ Molar mass of $CaCO_3 = 100 \ g/mol$ Molar mass of $CaO = 56 \ g/mol$ From the stoichiometry,$56 \ g$ of $CaO$ is produced from $100 \ g$ of pure $CaCO_3$. Since $56 \ g$ of $CaO$ is actually produced,the amount of pure $CaCO_3$ present in the sample is $100 \ g$. Percentage Purity $= \frac{	ext{Mass of pure } CaCO_3}{	ext{Mass of impure sample}} 	imes 100 \% = \frac{100 \ g}{150 \ g} 	imes 100 \% = 66.67 \%$

$150 \ g$ $CaCO_3$ sample was taken. On its complete decomposition $56 \ g$ $CaO$ is produced. The $\%$ purity of the sample is .............. $\%$

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