Đáp án:
\(\begin{array}{l}
b)\\
\% {m_{Al}} = 13,5\% \\
\% {m_{Cu}} = 86,5\% \\
c)\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 17,1g\\
d)\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = 0,15l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = 0,15 \times \frac{2}{3} = 0,1\,mol\\
\% {m_{Al}} = \dfrac{{0,1 \times 27}}{{20}} \times 100\% = 13,5\% \\
\% {m_{Cu}} = 100 - 13,5 = 86,5\% \\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,05 \times 342 = 17,1g\\
d)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,15\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,15}}{1} = 0,15l
\end{array}\)