When a mixture of 4.44 , g of CaCl_2 and NaCl | vedclass

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Question

(A) The decomposition reaction is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. From the stoichiometry,$100 \, g$ of $CaCO_3$ produces $56 \, g$ of $CaO$

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) The decomposition reaction is: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$. From the stoichiometry,$100 \, g$ of $CaCO_3$ produces $56 \, g$ of $CaO$. Therefore,$0.56 \, g$ of $CaO$ corresponds to $\frac{100}{56} 	imes 0.56 = 1 \, g$ of $CaCO_3$. The reaction between $CaCl_2$ and $Na_2CO_3$ is: $CaCl_2 + Na_2CO_3 ightarrow CaCO_3 + 2NaCl$. Since $111 \, g$ of $CaCl_2$ produces $100 \, g$ of $CaCO_3$,$1 \, g$ of $CaCO_3$ is produced by $\frac{111}{100} 	imes 1 = 1.11 \, g$ of $CaCl_2$. The mass of $NaCl$ in the mixture is $4.44 \, g - 1.11 \, g = 3.33 \, g$. The percentage of $NaCl$ is $\frac{3.33}{4.44} 	imes 100 = 75 \%$.

When a mixture of $4.44 \, g$ of $CaCl_2$ and $NaCl$ is treated with a sodium carbonate solution,$Ca^{2+}$ ions precipitate as $CaCO_3$. Upon heating,this precipitate yields $0.56 \, g$ of $CaO$. Calculate the percentage of $NaCl$ in the original mixture. (Atomic mass of $Ca = 40 \, g/mol$)

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