\(n_{H_2S}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{NaOH}=0,15.1=0,15\left(mol\right)\)
Xét \(T=\dfrac{n_{NaOH}}{n_{H_2S}}=\dfrac{0,15}{0,1}=1,5\) => Tạo cả 2 muối \(Na_2S,NaHS\)
PTHH: \(2NaOH+H_2S\rightarrow Na_2S+2H_2O\)
0,15---->0,075---->0,075
\(\rightarrow n_{H_2S\left(dư\right)}=0,1-0,075=0,025\left(mol\right)\)
PTHH: \(Na_2S+H_2S\rightarrow2NaHS\)
0,025<--0,025--->0,05
Muối thu được gồm \(\left\{{}\begin{matrix}Na_2S:n_{Na_2S}=0,075-0,025=0,05\left(mol\right)\\NaHS:n_{NaHS}=0,05\left(mol\right)\end{matrix}\right.\)
\(\rightarrow m_{muối.X}=0,05.78+0,05.56=6,7\left(g\right)\)
\(n_{H_2S}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(V_{NaOH}=150ml=0,15\left(l\right)\)
\(n_{NaOH}=CM.V=1.0,15=0,15\left(mol\right)\)
\(H_2S+NaOH\rightarrow Na_2S+H_2O\)
\(m_{Na_2S}=0,1.78=7,8\left(g\right)\)
\(m_{NaOH}=\left(0,15-0,1\right).40=2\left(g\right)\)
