Đáp án:
b) 16,92 g
c) 21,6 g
d) 31,36%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
nS{O_2} = \dfrac{{3,584}}{{22,4}} = 0,16\,mol\\
hh:Cu(a\,mol),Fe(b\,mol)\\
64a + 56b = 6,24\\
a + 1,5b = 0,16\\
\Rightarrow a = 0,01;b = 0,1\\
\% mCu = \dfrac{{0,01 \times 64}}{{6,24}} \times 100\% = 10,26\% \\
\% mFe = 100 - 10,26 = 89,74\% \\
b)\\
nCa{(OH)_2} = 0,1 \times 1 = 0,1\,mol\\
T = \dfrac{{0,1}}{{0,16}} = 0,625 \Rightarrow \\
Ca{(OH)_2} + S{O_2} \to CaS{O_3} + {H_2}O(1)\\
CaS{O_3} + S{O_2} + {H_2}O \to Ca{(HS{O_3})_2}(2)\\
nCaS{O_3}(1) = nS{O_2}(1) = 0,1\,mol\\
nCaS{O_3}(2) = nS{O_2}(2) = 0,16 - 0,1 = 0,06\,mol\\
nCaS{O_3} = 0,1 - 0,06 = 0,04\,mol\\
nCa{(HS{O_3})_2} = 0,06\,mol\\
\Rightarrow m = mCaS{O_3} + mCa{(HS{O_3})_2} = 16,92g\\
c)\\
m = mF{e_2}{(S{O_4})_3} + mCuS{O_4} = 21,6g\\
d)\\
n{H_2}S{O_4} = 2nS{O_2} = 0,32\,mol\\
C\% {H_2}S{O_4} = \dfrac{{0,32 \times 98}}{{100}} \times 100\% = 31,36\%
\end{array}\)
