a,

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$Mg+H_2SO_4\to MgSO_4+H_2$

$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2$

b,

$n_{Mg}=\dfrac{0,6}{24}=0,025(mol)$

$n_{H_2}=\dfrac{1,568}{22,4}=0,07(mol)$

Theo PTHH:

$1,5n_{Al}+n_{Mg}=n_{H_2}$

$\Rightarrow n_{Al}=0,03$

$\%m_{Mg}=\dfrac{0,6.100}{0,6+0,03.27}=42,6\%$

$\Rightarrow \%m_{Al}=57,4\%$

Đáp án:

 57,45% , 42,55%

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
b)\\
{m_{Mg}} = m = 0,6g\\
{n_{Mg}} = \dfrac{{0,6}}{{24}} = 0,025\,mol\\
{n_{{H_2}}} = \dfrac{{1,568}}{{22,4}} = 0,07\,mol\\
{n_{Al}} = \dfrac{{(0,07 - 0,025) \times 2}}{3} = 0,03\,mol\\
\% {m_{Al}} = \dfrac{{0,03 \times 27}}{{0,03 \times 27 + 0,6}} \times 100\%  = 57,45\% \\
\% {m_{Mg}} = 100 - 57,45 = 42,55\% 
\end{array}\)