Ta có:
\(n_{H2S}=\frac{7,168}{22,4}=0,32\left(mol\right)\)
\(m_{NaOH}=8.10\%=0,8\left(g\right)\Rightarrow n_{NaOH}=\frac{0,8}{40}=0,02\left(mol\right)\)
\(\frac{n_{NaOH}}{n_{H2S}}=\frac{0,02}{0,32}< 1\)
Nên H2S dư
\(NaOH+H_2S\rightarrow NaHS+H_2O\)
\(n_{NaHS}=n_{NaOH}=0,02\left(mol\right)\)
\(\Rightarrow m_{NaHS}=0,02.\left(23+1+32\right)=1,12\left(g\right)\)