Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{CuO}} = 10,75\% \\
\% {m_{Fe}} = 67,74\% \\
\% {m_{MgO}} = 21,51\% \\
b)\\
{C_\% }FeS{O_4} = 11,69\% \\
{C_\% }CuS{O_4} = 1,37\% \\
{C_\% }MgS{O_4} = 4,1\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
T{N_1}:\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
T{N_2}:\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}(T{N_1})}} = \dfrac{{0,224}}{{22,4}} = 0,01\,mol\\
{n_{CuO}} = {n_{{H_2}(T{N_1})}} = 0,01\,mol\\
{n_{{H_2}(T{N_2})}} = \dfrac{{2,016}}{{22,4}} = 0,09\,mol\\
{n_{Fe}} = {n_{{H_2}(T{N_2})}} = 0,09\,mol\\
\% {m_{CuO}} = \dfrac{{0,01 \times 80}}{{7,44}} \times 100\% = 10,75\% \\
\% {m_{Fe}} = \dfrac{{0,09 \times 56}}{{7,44}} \times 100\% = 67,74\% \\
\% {m_{MgO}} = 100 - 10,75 - 67,74 = 21,51\% \\
b)\\
{m_{MgO}} = 7,44 - 0,01 \times 80 - 0,09 \times 56 = 1,6g\\
{n_{MgO}} = \dfrac{{1,6}}{{40}} = 0,04\,mol\\
{n_{{H_2}S{O_4}}} = 0,04 + 0,01 + 0,09 = 0,14\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,14 \times 98}}{{12,5\% }} = 109,76g\\
{m_{{\rm{dd}}spu}} = 7,44 + 109,76 - 0,09 \times 2 = 117,02g\\
{C_\% }FeS{O_4} = \dfrac{{0,09 \times 152}}{{117,02}} \times 100\% = 11,69\% \\
{C_\% }CuS{O_4} = \dfrac{{0,01 \times 160}}{{117,02}} \times 100\% = 1,37\% \\
{C_\% }MgS{O_4} = \dfrac{{0,04 \times 120}}{{117,02}} \times 100\% = 4,1\%
\end{array}\)
