A mixture of CaCl_{2} and NaCl weighing 4.44 | vedclass

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(A) The chemical reactions involved are:$CaCl_{2} + Na_{2}CO_{3} \rightarrow CaCO_{3} + 2NaCl$$CaCO_{3} \xrightarrow{\Delta} CaO + CO_{2}$From the sto

Vedclass · Accepted Answer

$75$

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(A) The chemical reactions involved are:$CaCl_{2} + Na_{2}CO_{3} ightarrow CaCO_{3} + 2NaCl$$CaCO_{3} \xrightarrow{\Delta} CaO + CO_{2}$From the stoichiometry,$1 \ mol$ of $CaO$ is obtained from $1 \ mol$ of $CaCO_{3}$,which in turn comes from $1 \ mol$ of $CaCl_{2}$.Molar mass of $CaO = 40 + 16 = 56 \ g/mol$.Moles of $CaO = \frac{0.56 \ g}{56 \ g/mol} = 0.01 \ mol$.Therefore,moles of $CaCl_{2} = 0.01 \ mol$.Molar mass of $CaCl_{2} = 40 + (2 	imes 35.5) = 111 \ g/mol$.Mass of $CaCl_{2} = 0.01 \ mol 	imes 111 \ g/mol = 1.11 \ g$.Mass of $NaCl$ in the mixture $= 4.44 \ g - 1.11 \ g = 3.33 \ g$.Percentage of $NaCl = \frac{3.33 \ g}{4.44 \ g} 	imes 100 = 75 \%$.

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