`n_{H_2S}=\frac{2,24}{22,4}=0,1\ (mol)`
`n_{NaOH}=0,2.0,7=0,14\ (mol)`
Ta có: `T=\frac{n_{NaOH}}{n_{H_2S}}=\frac{0,14}{0,1}=1,4`
`=>1<T<2`
`=>` Tạo hai muối `NaHS` và `Na_2S`
`H_2S+NaOH\to NaHS+H_2O`
`H_2S+2NaOH\to Na_2S+H_2O`
Đặt: `n_{NaHS}=x\ (mol);\ n_{Na_2S}=y\ (mol)`
Ta có hệ: $\begin{cases}x+y=0,1\\x+2y=0,14\end{cases}$`=>`$\begin{cases}x=0,06\ (mol)\\y=0,04\ (mol)\end{cases}$
`=>m_{\text{muối}}=m_{NaHS}+m_{Na_2S}`
`=>m_{\text{muối}}=0,06.56+0,04.78=6,48\ (g)`