\(Fe + H_2SO_4 \to FeSO_4 + H_2\)

Chất rắn A : Cu

Theo PTHH : 

\(n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ \Rightarrow m_{Fe} = 0,15.56 = 8,4(gam)\)

Suy ra : \(m_{Cu} = 1,48 - 8,4 = -6,92 < 0\)→ Sai đề

\(Đặt:n_{Al}=u\left(mol\right);n_{Fe}=v\left(mol\right)\left(u,v>0\right)\\ n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56u=22\\1,5a+u=0,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\\u=0,2\end{matrix}\right.\\ \Rightarrow m_{Al}=0,4.27=10,8\left(g\right);m_{Fe}=56.0,2=11,2\left(g\right)\\ n_{HCl}=2.0,8=1,6\left(mol\right)\\ m_{HCl}=1,6.36,5=58,4\left(g\right)\\ m_{ddHCl}=\dfrac{58,4.100}{25}=233,6\left(g\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

x           3x             x             1,5x

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

y         2y              y          y

\(\left\{{}\begin{matrix}27x+56y=22\\1,5x+y=\dfrac{17,92}{22,4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

\(m_{Al}=0,4\cdot27=10,8g\)

\(m_{Fe}=22-10,8=11,2g\)

\(m_{HCl}=36,5\cdot\left(3x+2y\right)=36,5\cdot\left(3\cdot0,4+2\cdot0,2\right)=58,4g\)

\(m_{ddHCl}=\dfrac{m_{HCl}}{C\%}\cdot100\%=\dfrac{58,4}{25\%}\cdot100\%=233,6g\)

Theo gt ta có: $n_{H_2}=0,8(mol)$

a, Gọi số mol Fe và Al lần lượt là a;b(mol)

$Fe+2HCl\rightarrow FeCl_2+H_2$

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

Ta có: $56a+27b=22;a+1,5b=0,8$

Giải hệ ta được $a=0,2;b=0,4$

Do đó $\%m_{Fe}=50,9\%;\%m_{Al}=49,1\%$

b, Sau phản ứng dung dịch chứa 0,2mol $FeCl_2$ và 0,4mol $AlCl_3$

$m_{dd}=22+1,6.36,5:7,3\%-0,8.2=820,4(g)$

Do đó $\%C_{FeCl_2}=3,09\%;\%C_{AlCl_3}=6,5\%$

$a\bigg)$

Đặt $n_{Al}=x(mol);n_{Fe}=y(mol)$

$\to 27x+56y=22(1)$

BTe: $1,5x+y=n_{H_2}=\dfrac{17,92}{22,4}=0,8(2)$

Từ $(1)(2)\to x=0,4(mol);y=0,2(mol)$

$\to \%m_{Al}=\dfrac{0,4.27}{22}.100\%\approx 49,09\%$

$\to \%m_{Fe}=100-49,09=50,91\%$

$b\bigg)$

Bảo toàn H: $n_{HCl}=2n_{H_2}=1,6(mol)$

$\to m_{dd_{HCl}}=\dfrac{1,6.36,5}{25\%}=233,6(g)$

$\to m_{dd\, sau}=22+233,6-0,8.2=254(g)$

Bảo toàn Al,Fe: $n_{AlCl_3}=0,4(mol);n_{FeCl_2}=0,2(mol)$

$\to \begin{cases} C\%_{AlCl_3}=\dfrac{0,4.133,5}{254}.100\%\approx 21,02\%\\ C\%_{FeCl_2}=\dfrac{0,2.127}{254}.100\%=10\% \end{cases}$

a)

\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂ (1)

             0,6<----------------------0,3

=> m_Na = 0,6.23 = 13,8 (g)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1<-0,2

=> m_Fe = 0,1.56 = 5,6 (g)

m_Cu = 10 (g)

\(\left\{{}\begin{matrix}\%Na=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%Fe=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%Cu=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

             \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=\dfrac{17,4}{\dfrac{0,3}{y}}=58y\left(g/mol\right)\)

=> 56x = 42y

=> \(\dfrac{x}{y}=\dfrac{3}{4}\) => CTHH: Fe₃O₄

a)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

            0,6<----------------------0,3

             Fe + 2HCl --> FeCl₂ + H₂

            0,1<--0,2

=> \(\left\{{}\begin{matrix}m_{Na}=0,6.23=13,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=10\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%m_{Fe}=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%m_{Cu}=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)
PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

            \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=56x+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\left(g/mol\right)\)

=> \(\dfrac{x}{y}=\dfrac{3}{4}\)

=> CTHH: Fe₃O₄