Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{CuO}} = 50\% \\
\% {m_{F{e_2}{O_3}}} = 50\% \\
b)\\
{C_\% }CuS{O_4} = 5,06\% \\
{C_\% }F{e_2}{(S{O_4})_3} = 6,33\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
hh:CuO(a\,mol),F{e_2}{O_3}(b\,mol)\\
{n_{{H_2}S{O_4}}} = 0,5 \times 1 = 0,5\,mol\\
80a + 160b = 32\\
a + 3b = 0,5\\
\Rightarrow a = 0,2;b = 0,1\\
\% {m_{CuO}} = \dfrac{{0,2 \times 80}}{{32}} \times 100\% = 50\% \\
\% {m_{F{e_2}{O_3}}} = 100 - 50 = 50\% \\
b)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 500 \times 1,2 = 600g\\
{m_{{\rm{dd}}spu}} = 32 + 600 = 632g\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,2\,mol\\
{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,1\,mol\\
{C_\% }CuS{O_4} = \dfrac{{0,2 \times 160}}{{632}} \times 100\% = 5,06\% \\
{C_\% }F{e_2}{(S{O_4})_3} = \dfrac{{0,1 \times 400}}{{632}} \times 100\% = 6,33\%
\end{array}\)
