Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{HCl}} = 2{n_{Fe}} = 0,4\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,4 \times 36,5}}{{15\% }} = 97,33g\\
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 11,2 + 97,33 - 0,2 \times 2 = 108,13g\\
{C_\% }FeC{l_2} = \dfrac{{0,2 \times 127}}{{108,13}} \times 100\% = 23,49\% \\
2)\\
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15\,mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{Mg}} = 0,15\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,15}}{2} = 0,075l = 75\,ml\\
c)\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,15\,mol\\
{C_M}MgS{O_4} = \dfrac{{0,15}}{{0,075}} = 2M
\end{array}\)
