A mixture of CaCl_2 and NaCl weighing 4.44 g | vedclass

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(A) The chemical reactions involved are:$CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 + 2NaCl$$CaCO_3 \xrightarrow{\Delta} CaO + CO_2$From the stoichiometry,$

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The chemical reactions involved are:$CaCl_2 + Na_2CO_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 $CaO = 40 + 16 = 56 \ g/mol$.Molar mass of $CaCl_2 = 40 + 2 	imes 35.5 = 111 \ g/mol$.Given mass of $CaO = 0.56 \ g$,which is $0.56 / 56 = 0.01 \ mol$.Therefore,moles of $CaCl_2 = 0.01 \ mol$.Mass of $CaCl_2 = 0.01 	imes 111 = 1.11 \ g$.Mass of $NaCl$ in the mixture = Total mass - Mass of $CaCl_2 = 4.44 \ g - 1.11 \ g = 3.33 \ g$.Percentage of $NaCl = (3.33 / 4.44) 	imes 100 = 75\%$.

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