Gọi \(\left\{{}\begin{matrix}n_{H2S}:x\left(mol\right)\\n_{O2}:y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{hh}=x+y=\frac{20,16}{22,4}=0,9\left(mol\right)\)
\(M_{hh}=16,22.M_{H2}=16,22.2=32,44\)
\(\Rightarrow m_{hh}=34x+32y=32,44.0,9=29,196\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,198\\y=0,702\end{matrix}\right.\)
Vì % số mol=% thể tích
\(\%V_{H2S}=\frac{0,198}{0,9}.100\%=22\%\Rightarrow\%V_{H2}=78\%\)
Đốt cháy hỗn hợp trên thì chỉ có H2S cháy
\(2H_2S+3O_2\rightarrow2H_2O+2SO_2\)
\(\Leftrightarrow n_{H2O}=n_{SO2}=n_{H2S}=0,198\left(mol\right)\)
\(\Rightarrow m_{H2O}=0,198.18=3,564\left(g\right)\)
\(\Rightarrow V_{H2O}=3,564\left(ml\right)\)
\(SO_2+H_2O\rightarrow H_2SO_3\)
\(n_{H2SO3}=n_{SO2}=0,198\left(mol\right)\)
\(V_{dd}=94,6+3,564=98,164\left(ml\right)=0,098\left(ml\right)\)
\(\Rightarrow CM_{H2SO3}=\frac{0,198}{0,098}=2M\)
\(m_{dd\left(spu\right)}=m_{SO2}+m_{H2O}+m_{H2O\left(bđ\right)}=0,198.64+3,564+94,6.1=110,836\left(g\right)\)
\(\Rightarrow C\%_{H2SO3}=\frac{0,198.82}{110,836}=14,65\%\)
