Đáp án + Giải thích các bước giải:

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

$b\big)$

$n_{H_2SO_4}=\frac{200\times 1,47\%}{98}=0,03(mol)$

Theo PT: $n_{Al}=\frac{2}{3}\times n_{H_2SO_4}=0,02(mol)$

$\to m_{Al}=0,02\times 27=0,54(g)$

$c\big)$

Theo PT: $n_{Al_2(SO_4)_3}=\frac{1}{2}\times n_{Al}=0,01(mol)$

$\to m_{Al_2(SO_4)_3}=0,01\times 342=3,42(g)$

$d\big)$

Theo PT: $n_{H_2}=n_{H_2SO_4}=0,03(mol)$

$\to V_{H_2(đktc)}=0,03\times 22,4=0,672(l)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
b)\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 1,47\% }}{{98}} = 0,03mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}S{O_4}}} = 0,02mol\\
{m_{Al}} = 0,02 \times 27 = 0,54g\\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{3} = 0,01mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,01 \times 342 = 3,42g\\
d)\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,03mol\\
{V_{{H_2}}} = 0,03 \times 22,4 = 0,672l
\end{array}\)