Đáp án:
30,4 g
Giải thích các bước giải:
\(\begin{array}{l}
hh:FeO(a\,mol),CuO(b\,mol)\\
P1:\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
0,5a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5a\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
0,5b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5b\\
P2:\\
2FeO + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 4{H_2}O\\
0,5a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,25a\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
0,5b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5b\\
127 \times 0,5a + 135 \times 0,5b = 26,2\\
400 \times 0,25a + 160 \times 0,5b = 36\\
\Rightarrow a = b = 0,2\,mol\\
m = mFeO + mCuO = 0,2 \times 72 + 0,2 \times 80 = 30,4g
\end{array}\)
