\(\begin{array}{l}
12)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
CuS + 2HCl \to CuC{l_2} + {H_2}S\\
nhh = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
hh:Mg(a\,mol),CuS(b\,mol)\\
24a + 96b = 7,2\\
a + b = 0,15\\
\Rightarrow a = 0,1;b = 0,05\\
\% mMg = \dfrac{{0,1 \times 24}}{{7,2}} \times 100\% = 33,33\% \\
\% mCuS = 100 - 33,33 = 66,67\% \\
b)\\
n{H_2}S = nCuS = 0,05\,mol\\
n{H_2} = nMg = 0,1\,mol\\
Mhh = \dfrac{{0,1 \times 2 + 0,05 \times 32}}{{0,15}} = 12g/mol\\
dM/{O_2} = \frac{{12}}{{32}} = 0,375\\
13)\\
a)\\
nhh = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
Mg + S \to MgS\\
MgS + 2HCl \to MgC{l_2} + {H_2}S\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
hh:Mg(a\,mol),S(b\,mol)\\
n{H_2}S = nMgS = nS = b\,mol\\
n{H_2} = nMg = a - b\,mol\\
a - b + b = 0,2\\
24a + 32b = 8\\
\Rightarrow a = 0,2;b = 0,1\\
\% mMg = \dfrac{{0,2 \times 24}}{8} \times 100\% = 60\% \\
\% mS = 100 - 60 = 40\% \\
b)\\
MB = \dfrac{{0,1 \times 2 + 0,1 \times 34}}{{0,2}} = 18g/mol\\
dMB/{H_2} = \dfrac{{18}}{2} = 9\\
c)\\
nNaOH = 0,075 \times 2 = 0,15\,mol\\
T = \dfrac{{nNaOH}}{{n{H_2}S}} = \dfrac{{0,15}}{{0,1}} = 1,5\\
1 < T < 1,5\\
\Rightarrow PTHH:\\
2NaOH + {H_2}S \to N{a_2}S + 2{H_2}O(1)\\
N{a_2}S + {H_2}S \to 2NaHS(2)\\
nN{a_2}S(1) = n{H_2}S(1) = 0,075\,mol\\
nN{a_2}S(2) = n{H_2}S(2) = 0,1 - 0,075 = 0,025\,mol\\
nN{a_2}S = 0,075 - 0,025 = 0,05\,mol\\
nNaHS = 2 \times 0,025 = 0,05\,mol\\
mN{a_2}S = 0,05 \times 78 = 3,9g\\
mNaHS = 0,05 \times 56 = 2,8g
\end{array}\)
