2Al+3H2SO4→Al2(SO4)3+3H2

         0,1         0,1/3             0,1

nH2=0,3*3/2=0,45mol

nAl=8,1/27=0,3mol

2H2+O2→2H2O

nkk=5,6/22,4=0,25mol

⇒nO2=0,25/5=0,05mol

⇒nH2=0,1mol

 H=0,1/0,45*100%=22,22%

Vdd H2SO4=0,1/0,05=2l

Vdd sau=V dd H2SO4=2l

CM Al2(SO4)3=0,1/3/2=1/60M

Em tham khảo nha :

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
2{H_2} + {O_2} \to 2{H_2}O\\
{V_{{O_2}}} = \dfrac{{{V_{kk}}}}{5} = 1,12l\\
{n_{{O_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{{H_2}}} = 2{n_{{O_2}}} = 0,1mol\\
{n_{Al}} = \dfrac{{8,1}}{{27}} = 0,3mol\\
{n_{{H_2}(1)}} = \dfrac{3}{2}{n_{Al}} = 0,45mol\\
H = \dfrac{{0,1}}{{0,45}} \times 100\%  = 22,22\% \\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,15mol\\
{n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 0,45mol\\
{V_{{H_2}S{O_4}}} = \dfrac{{0,45}}{{0,05}} = 9l\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,15}}{9} = \frac{1}{{60}}M
\end{array}\)