Em tham khảo nha :

\(\begin{array}{l}
2)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,05mol\\
{m_{Fe}} = 0,05 \times 56 = 2,8g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,1mol\\
{m_{HCl}} = 0,1 \times 36,5 = 3,65g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{3,65 \times 100}}{{10}} = 36,5g\\
3)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
{m_{{H_2}S{O_4}}} = 0,3 \times 98 = 29,4g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{29,4 \times 100}}{{10}} = 294g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} 2.\\ a)\\ n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\Fe+2HCl\to FeCl_2+H_2\\n_{Fe}=n_{H_2}=0,05(mol)\\m_{Fe}=0,05×56=2,8(g)\\b)\\n_{HCl}=2n_{H_2}=0,1(mol)\\m_{dd\ HCl}=\dfrac{0,1×36,5}{10\%}=36,5(g)\\3.\\a)\\n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)\\m_{Al}=0,2×27=5,4(g)\\b)\\n_{H_2SO_4}=n_{H_2}=0,3(mol)\\m_{dd\ H_2SO_4}=\dfrac{0,3×98}{10\%}=294(g)\end{array}\)