PTHH:\(Na_2SO_3+CaCl_2\rightarrow2NaCl+CaSO_3\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_3}=\dfrac{265\cdot10\%}{126}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2}=\dfrac{500\cdot6,66\%}{111}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỷ số: \(\dfrac{53}{252}< \dfrac{0,3}{1}\) \(\Rightarrow\) CaCl2 còn dư, Na2SO3 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=\dfrac{53}{126}\left(mol\right)\\n_{CaSO_3}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=\dfrac{53}{126}\cdot58,5\approx24,61\left(g\right)\\m_{CaSO_3}=\dfrac{53}{252}\cdot120\approx25,24\left(g\right)\\m_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\cdot111\approx9,95\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNa_2SO_3}+m_{ddCaCl_2}-m_{CaSO_3}=739,76\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{24,61}{739,76}\cdot100\%\approx3,33\%\\C\%_{CaCl_2\left(dư\right)}=\dfrac{9,95}{739,76}\cdot100\%\approx1,35\%\end{matrix}\right.\)
