\(T = \dfrac{{nNaOH}}{{n{H_2}S}}\)
T<1: tạo NaHS
1<T<2: tạo 2 muối NaHS và $Na_2S$
T>2: tạo $Na_2S$
\(\begin{array}{l}
1)\\
n{H_2}S = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
nNaOH = 0,2 \times 1 = 0,2\,mol\\
\dfrac{{nNaOH}}{{n{H_2}S}} = \dfrac{{0,2}}{{0,3}} = 0,67\\
NaOH + {H_2}S \to NaHS + {H_2}O\\
nNaHS = nNaOH = 0,2\,mol\\
mNaHS = 0,2 \times 56 = 11,2g\\
2)\\
n{H_2}S = \dfrac{{1,12}}{{22,4}} = 0,05\,mol\\
nNaOH = 0,2 \times 1 = 0,2\,mol\\
\dfrac{{nNaOH}}{{n{H_2}S}} = \dfrac{{0,2}}{{0,05}} = 4\\
2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O\\
nN{a_2}S = n{H_2}S = 0,05\,mol\\
mN{a_2}S = 0,05 \times 78 = 3,9g\\
3)\\
n{H_2}S = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
nNaOH = 0,2 \times 1 = 0,2\,mol\\
\dfrac{{nNaOH}}{{n{H_2}S}} = \dfrac{{0,2}}{{0,15}} = 1,33\\
2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O(1)\\
N{a_2}S + {H_2}S \to 2NaHS\\
nN{a_2}S(1) = n{H_2}S(1) = 0,1\,mol\\
nN{a_2}S(2) = n{H_2}S(2) = 0,15 - 0,1 = 0,05\,mol\\
nN{a_2}S = 0,1 - 0,05 = 0,05\,mol\\
nNaHS = 0,05 \times 2 = 0,1\,mol\\
m = mN{a_2}S + mNaHS = 9,5g
\end{array}\)
