Gọi `x = n_(Al)   ;  y = n_(Fe) `

`n_(H_2) = 1.12/ 22.4  = 0.05 ( mol )`

`PTHH :  2Al  +   3H_2SO_4  ->  Al_2(SO_4)_3  + 3 H_2 ↑ `

               `  x                    -               x/2         -        3/2x`

             ` Fe    +  H_2SO_4    ->  FeSO_4 +  H_2`

               `  y              -                y            -       y` 

Ta có hệ pt : $\left \{ {{27x+56y=1.66} \atop {3/2x+y=0.05}} \right.$ `=>`$\left \{ {{x = 0.02} \atop {y=0.02}} \right.$ 

`-> %_(Al) =`$\dfrac{0.02×27}{1.66}$`×100% = 32.53%` 

`-> %_(Fe) = 100% - 32.53% = 67.47%`

b) 

`n_(Al_2(SO_4)_3) = 0.02 / 2 = 0.01 ( mol ) `

`-> m_(Al_2(SO_4)_3) = 0.01 × 342 = 3.42 ( g ) `

`n_(FeSO_4) = 0.02 mol `

`-> m_(FeSO_4) = 0.02 ×152 = 3.04 ( g ) ` 

c) `m_(H_2SO_4) = 0.05 × 98 = 4.9 ( g ) `

`m_(ddH_2SO_4) = V × D = 200 × 1.05 = 210 ( g ) `

`-> C%_(H_2SO_4) = 4.9 / 210 × 100% = 2.33%`

Đáp án:

\(\% {m_{Al}}  = 32,5\%; \% {m_{Fe}} = 67,5\% \)

\( C{\% _{{H_2}S{O_4}}}= 2,333\% \)

Giải thích các bước giải:

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

\( \to 27x + 56y = 1,66\)

Phản ứng xảy ra:

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

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol = }}\frac{3}{2}{n_{Al}} + {n_{Fe}} = 1,5x + y\)

Giải được:

\(x=0,02;y=0,03\)

\( \to {m_{Al}} = 0,02.27 = 0,54{\text{ gam}}\)

\( \to \% {m_{Al}} = \frac{{0,54}}{{1,66}} = 32,5\%  \to \% {m_{Fe}} = 67,5\% \)

\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,05{\text{ mol}}\)

\({m_{{H_2}S{O_4}}} = 0,05.98 = 4,9{\text{ gam}}\)

\({m_{dd\;{{\text{H}}_2}S{O_4}}} = 200.1,05 = 210{\text{ gam}}\)

\( \to C{\% _{{H_2}S{O_4}}} = \frac{{4,9}}{{210}} = 2,333\% \)