Đáp án:

 `a, 2,7gam ; 17,1 gam`

`b, 14,7gam`

Giải thích các bước giải:

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

`n_{H_2}=(3,6)/(24)=0,15(mol)`

Theo pthh:

`n_{Al}=2/3 .n_{H_2}=2/3 .0,15=0,1(mol)`

`-> m_{Al}=0,1.27=2,7(gam)`

`<=> a=2,7gam`

`n_{Al_2(SO_4)_3}=1/3 .n_{H_2}=1/3 .0,15=0,05(mol)`

`-> m_{Al_2(SO_4)_3}=0,05.342=17,1(gam)`

`<=> b=17,1gam`

`b,`

Theo pthh:

`n_{H_2SO_4}=n_{H_2}=0,15(mol)`

`-> m_{H_2SO_4}=0,15.98=14,7(gam)`

Đáp án:

\( a = 2,7{\text{ gam}}\)

\(b = 17,1{\text{ gam}}\)

\({m_{{H_2}S{O_4}}} = 14,7{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

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

Ta có:

\({n_{{H_2}}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol}}\)

Theo phản ứng:

\({n_{Al}} = \frac{2}{3}{n_{{H_2}}} = 0,1{\text{ mol;}}{{\text{n}}_{A{l_2}{{(S{O_4})}_3}}} = \frac{1}{3}{n_{{H_2}}} = 0,0{\text{ mol}}\)

\( \to a = {m_{Al}} = 0,1.27 = 2,7{\text{ gam}}\)

\(b = {m_{A{l_2}{{(S{O_4})}_3}}} = 0,05.(27.2 + 96.3) = 17,1{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,15{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,15.98 = 14,7{\text{ gam}}\)