Đáp án:

\({V_{{H_2}}} = 11,2{\text{ lít}}\)

\(\% {m_{Zn}} = 95\%; \% {m_{Al}} = 5\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

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

Ta có:

\({n_{HCl}} = \frac{{16,425}}{{36,5}} = 0,45{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{26,95}}{{98}} = 0,275{\text{ mol}}\)

\( \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} + {n_{{H_2}S{O_4}}} = \frac{1}{2}.0,45 + 0,275 = 0,5{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,5.22,4 = 11,2{\text{ lít}}\)

Gọi số mol \(Zn;Al\) lần lượt là \(x;y\)

\( \to 65x + 27y = 28,7{\text{ gam}}\)

\({n_{{H_2}}} = {n_{Zn}} + \frac{3}{2}{n_{Al}} = x + 1,5y = 0,5\)

Giải được:

\(x = \frac{{197}}{{470}};y = \frac{{38}}{{705}}\)

\( \to {m_{Zn}} = 65x = 27,25{\text{ gam}}\)

\( \to \% {m_{Zn}} = \frac{{27,25}}{{28,7}} = 95\%  \to \% {m_{Al}} = 5\% \)

$n_{HCl}=\dfrac{16,425}{36,5}=0,45(mol)$

$n_{H_2SO_4}=\dfrac{26,95}{98}=0,275(mol)$

$\to n_H=n_{HCl}+2n_{H_2SO_4}=1(mol)$

Gọi chung hai axit là $HX$ ($1$ mol)

Gọi $x$, $y$ là số mol $Zn$, $Al$ 

$\to 65x+27y=28,7$

$Zn+2HX\to ZnX_2+H_2$

$2Al+6HX\to 2AlX_3+3H_2$

$\to 2x+3y=1$                       

Giải hệ ta có: $x=\dfrac{197}{470}; y=\dfrac{38}{705}$

$\%m_{Zn}=\dfrac{65x.100}{28,7}=94,9\%$

$\%m_{Al}=5,1\%$

Theo PTHH: $n_{H_2}=\dfrac{n_H}{2}=0,5(mol)$

$\to V_{H_2}=0,5.22,4=11,2l$