\(n_{H_2S}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2H2S + 3O2 --to--> 2SO2 + 2H2O
0,4--------------->0,4
\(m_{dd.NaOH}=50.1,28=64\left(g\right)\)
=> mNaOH = 64.25% = 16 (g)
=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
Xét \(\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,4}{0,4}=1\) => Thu được muối NaHSO3
PTHH: NaOH + SO2 --> NaHSO3
0,4------------->0,4
=> \(C\%_{NaHSO_3}=\dfrac{0,4.104}{0,4.64+64}.100\%=46,429\%\)