PTHH: \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\)
Ta có: \(n_{SO_2}=\frac{5,6}{22,4}=0,25\left(mol\right)\) \(\Rightarrow n_{Na_2SO_3}=n_{H_2SO_4}=0,25mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_3\left(LT\right)}=0,25\cdot126=31,5\left(g\right)\\m_{H_2SO_4\left(LT\right)}=0,25\cdot98\cdot62\%=15,19\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_3\left(Thực\right)}=\frac{31,5}{80\%}=39,375\left(g\right)\\m_{H_2SO_4\left(Thực\right)}=\frac{15,19}{80\%}=18,9875\left(g\right)\end{matrix}\right.\)