Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 34,62\% \\
\% {m_{A{l_2}{O_3}}} = 65,38\% \\
b)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 660g\\
c)\\
{C_\% }A{l_2}{(S{O_4})_3} = 10,13\% \\
{C_\% }{H_2}S{O_4} = 0,8711\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{Al}} = 0,3 \times \frac{2}{3} = 0,2\,mol\\
\% {m_{Al}} = \dfrac{{0,2 \times 27}}{{15,6}} \times 100\% = 34,62\% \\
\% {m_{A{l_2}{O_3}}} = 100 - 34,62 = 65,38\% \\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{{15,6 - 0,2 \times 27}}{{102}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} \text{ phản ứng } = 0,3 + 0,1 \times 3 = 0,6\,mol\\
{n_{{H_2}S{O_4}}} \text{ ban đầu }= 0,6 + 0,6 \times 10\% = 0,66\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 0,66 \times \dfrac{{98}}{{9,8\% }} = 660g\\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,2}}{2} + 0,1 = 0,2\,mol\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,6 \times 10\% = 0,06\,mol\\
{m_{{\rm{ddA}}}} = 15,6 + 660 - 0,3 \times 2 = 675g\\
{C_\% }A{l_2}{(S{O_4})_3} = \dfrac{{0,2 \times 342}}{{675}} \times 100\% = 10,13\% \\
{C_\% }{H_2}S{O_4} \text{ dư } = \dfrac{{0,06 \times 98}}{{675}} \times 100\% = 0,8711\%
\end{array}\)
