Đáp án:

Giải thích các bước giải:

$a)$ Đặt $n_{Fe}=x(mol), n_{Mg}=y(mol)$

$\to 56x+24y=8(g)\ (1)$

Bảo toàn electron:

$x+y=n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\ (2)$

Từ (1)(2), ta có:

$x=y=0,1(mol)$

$\%m_{Fe}=\dfrac{0,1.56}{8}=70\%$

$\%m_{Mg}=100\%-70\%=30\%$

$b)$ Bảo toàn $H:$

$n_{H_2SO_4}=n_{H_2}=0,2(mol)$

$C_{M_{H_2SO_4}}=\dfrac{0,2}{0,5}=0,4M$

Bảo toàn nguyên tố:

$n_{FeSO_4}=n_{Fe}=0,1(mol)$

$n_{MgSO_4}=n_{Mg}=0,1(mol)$

$C_{M_{FeSO_4}}=C_{M_{MgSO_4}}=\dfrac{0,1}{0,5}=0,2M$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
56a + 24b = 8
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,1\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{8} \times 100\%  = 70\% \\
\% Mg = 100 - 70 = 30\% \\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,2mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,2}}{{0,5}} = 0,4M\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,1mol\\
{C_{{M_{FeS{O_4}}}}} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
{C_{{M_{MgS{O_4}}}}} = \dfrac{{0,1}}{{0,5}} = 0,2M
\end{array}\)