Bài 1. hòa tan 18,4g hỗn hợp Fe Mg Cu trong lượng dư dung dịch H2 SO4 2M vừa đủ Sau phản ứng thấy còn 8 gam chất rắn không tan và 6,72 lít khí ở điều

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Fe\,(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
56a + 24b = 10,4
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,2\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{18,4}} \times 100\%  = 30,4\% \\
\% Cu = \dfrac{8}{{18,4}} \times 100\%  = 43,5\% \\
\% Mg = 100 – 43,5 – 30,4 = 26,1\% \\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3mol\\
{V_{{H_2}S{O_4}}} = \dfrac{{0,3}}{2} = 0,15l = 150ml\\
c)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Fe}} + {n_{Cu}} + {n_{Mg}} = 0,475mol\\
{V_{S{O_2}}} = 0,475 \times 22,4 = 10,64l\\
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,475mol\\
{m_{N{a_2}S{O_3}}} = 0,475 \times 126 = 59,85g\\
2)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
{m_{MgO}} = 7,6 – 3,6 = 4g\\
b)\\
{n_{MgS{O_4}}} = {n_{Mg}} + {m_{MgO}} = 0,15 + 0,1 = 0,25mol\\
{m_{MgS{O_4}}} = 0,25 \times 120 = 30g\\
c)\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
BaC{l_2} + MgS{O_4} \to BaS{O_4} + MgC{l_2}\\
{n_{BaS{O_4}}} = {n_{MgS{O_4}}} + {n_{{H_2}S{O_4}d}} = 0,3mol\\
{m_{BaS{O_4}}} = 0,3 \times 233 = 69,9g\\
d)\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
{n_{S{O_2}}} = {n_{Mg}} = 0,15mol\\
{V_{S{O_2}}} = 0,15 \times 22,4 = 3,36l
\end{array}\)