Đáp án:

 a) 90% và 10%

b) 800g

c) 8,45%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
n{H_2} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
 =  > nAl = 0,4\,mol\\
\% mAl = \dfrac{{0,4 \times 27}}{{12}} \times 100\%  = 90\% \\
\% mAg = 100 - 90 = 10\% \\
b)\\
n{H_2}S{O_4} = n{H_2} = 0,6\,mol\\
m{\rm{dd}}{H_2}S{O_4} = \dfrac{{0,6 \times 98}}{{7,35\% }} = 800g\\
c)\\
m{\rm{dd}}spu = 10,8 + 800 - 0,6 \times 2 = 809,6g\\
C\% A{l_2}{(S{O_4})_3} = \dfrac{{0,2 \times 342}}{{809,6}} \times 100\%  = 8,45\% 
\end{array}\)

Đáp án:

 a) \(\% {m_{Al}} = 90\% ;\% {m_{Ag}} = 10\% \)

b) \({m_{dd{{\text{H}}_2}S{O_4}}} = 800{\text{ gam}}\)

c) \(C{\% _{A{l_2}{{(S{O_4})}_3}}} = 8,45\% \)

Giải thích các bước giải:

 Chỉ có Al phản ứng, Ag không phản ứng.

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

\({n_{{H_2}}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol}} \to {{\text{n}}_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,4{\text{ mol}} \to {{\text{m}}_{Al}} = 0,4.27 = 10,8{\text{ gam}}\)

\(\to \% {m_{Al}} = \frac{{10,8}}{{12}} = 90\%  \to \% {m_{Ag}} = 10\% \)

Ta có: 

\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,6{\text{ mol}} \to {m_{{H_2}S{O_4}}} = 0,6.98 = 58,8{\text{ gam}} \to {{\text{m}}_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{58,8}}{{7,35\% }} = 800{\text{ gam}}\)

BTKL: \({m_{dd{\text{sau phản ứng}}}} = {m_{Al}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} - {m_{{H_2}}} = 10,8 + 800 - 0,6.2 = 809,6{\text{ gam}}\)

\({n_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Al}} = 0,2{\text{ mol}} \to {{\text{m}}_{A{l_2}{{(S{O_4})}_3}}} = 0,2.(27.2 + 96.3) = 68,4{\text{ gam}}\)

\(\to C{\% _{A{l_2}{{(S{O_4})}_3}}} = \frac{{68,4}}{{809,6}} = 8,45\% \)