\(n_{Na_2SO_3}=\dfrac{37,8}{126}=0,3\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{mdd\cdot C\%}{100\%}=\dfrac{200\cdot19,6\%}{100\%}=39,2\left(g\right)\)
--> \(n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
PTHH:
\(Na_2SO_3+H_2SO_4->Na_2SO_4+SO_2+H_2O\)
0,3 ---> 0,3 0,3 0,3 (mol)
So sánh số mol ta thấy \(n_{Na_2SO_3}< n_{H_2SO_4}\left(0,3< 0,4\right)\)
--> tính theo \(n_{Na_2SO_3}\)
\(V_{SO_2}=0,3\cdot22,4=6,72\left(l\right)\)
b) Sau phản ứng thu được \(Na_2SO_4,H_2SO_4dư\)
\(m_{ddsau}=m_{Na_2SO_3}+m_{ddH_2SO_4}-m_{SO_2}=37,8+200-64\cdot0,3=37,8+200-19,2=218,6\left(g\right)\)
\(n_{H_2SO_4dư}=0,4-0,3=0,1\left(mol\right)\) --> \(m_{H_2SO_4dư}=0,1\cdot98=9,8\left(g\right)\)
\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{42,6}{218,6}\cdot100\%\approx19,487\%\)
\(C\%_{H_2SO_4dư}=\dfrac{9,8}{218,6}\cdot100\%\approx4,483\%\)
