Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 58,33\% \\
\% {m_{CuO}} = 41,67\% \\
b)\\
{C_\% }CuS{O_4} = 18,66\% \\
{C_\% }F{e_2}{(S{O_4})_3} = 46,65\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{S{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Fe}} = \dfrac{{0,3 \times 2}}{3} = 0,2\,mol\\
\% {m_{Fe}} = \dfrac{{0,2 \times 56}}{{19,2}} \times 100\% = 58,33\% \\
\% {m_{CuO}} = 100 - 58,33 = 41,67\% \\
b)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{CuO}} = \dfrac{{19,2 - 0,2 \times 56}}{{80}} = 0,1\,mol\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = 0,3 \times 2 + 0,1 = 0,7\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,7 \times 98}}{{80\% }} = 85,75g\\
{m_{ddspu}} = 85,75 + 19,2 - 0,3 \times 64 = 85,75g\\
{C_\% }CuS{O_4} = \dfrac{{0,1 \times 160}}{{85,75}} \times 100\% = 18,66\% \\
{C_\% }F{e_2}{(S{O_4})_3} = \dfrac{{0,1 \times 400}}{{85,75}} \times 100\% = 46,65\%
\end{array}\)
