\(n_{Na_2SO_3}=\dfrac{12.6}{126}=0.1\left(mol\right)\)
\(n_{HCl}=\dfrac{800\cdot7.3\%}{100\%\cdot36.5}=1.6\left(mol\right)\)
\(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)
\(\text{Ban đầu : }\)\(0.1.............1.6\)
\(\text{Phản ứng:}\) \(0.1.........0.2..........0.2........0.1\)
\(\text{Kết thúc : }\) \(0...........1.4........0.2..........0.1\)
\(m_{dd}=12.6+800-0.1\cdot80=804.6\left(g\right)\)
\(C\%HCl\left(dư\right)=\dfrac{1.4\cdot36.5}{804.6}\cdot100\%=6.35\%\)
\(C\%NaCl=\dfrac{0.2\cdot58.5}{804.6}\cdot100\%=1.45\%\)
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