Em tham khảo nha :
\(\begin{array}{l}
2)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,3}}{{0,2}} = 1,5M\\
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{{H_2}}}}}{3} = 0,1mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,1 \times 342 = 34,2g\\
3)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
hh:Fe(a\,mol);Mg(b\,mol)\\
56a + 24b = 8(1)\\
a + b = 0,2(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,1\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
{m_{Mg}} = 8 - 5,6 = 2,4g\\
b){n_{HCl}} = 2{n_{Fe}} + 2{n_{Mg}} = 0,4mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,4}}{{0,2}} = 2M\\
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,1mol\\
{m_{MgC{l_2}}} = 0,1 \times 95 = 9,5g\\
{m_m} = 12,7 + 9,5 = 22,2g
\end{array}\)
