Đáp án D

\(n_{Zn}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(1\right)\)

\(m_{Muối}=m_{ZnCl_2}+m_{AlCl_3}=136a+133.5b=40.3\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.2\)

\(m_{hh}=0.1\cdot65+0.2\cdot27=11.9\left(g\right)\)

\(\%Zn=\dfrac{0.1\cdot65}{11.9}\cdot100\%=54.62\%\)

\(\%Al=100-54.62=45.38\%\)

Sửa đề: đktc → đkc

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Ta có: 24n_Mg + 56n_Fe = 13,2 (1)

\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=0,35\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{13,2}.100\%\approx36,36\%\\\%m_{Fe}\approx63,64\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)

⇒ m _{muối khan} = 0,2.95 + 0,15.127 = 38,05 (g)

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

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=0,2\cdot4=0,8mol\)

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

\(x\)   \(\rightarrow\)   \(3x\)            \(x\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 \(y\)   \(\rightarrow\) \(2y\)            \(y\)

\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)

\(\%m_{Zn}=100\%-45,38\%=54,62\%\)

b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)

\(V_{H_2}=0,4\cdot22.4=8,96l\)

a, Ta có: 24n_Mg + 56n_Fe = 12,8 (1)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,3\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

b, \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,8}{2}=0,4\left(l\right)\)

c, \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

Theo PT: \(\left\{{}\begin{matrix}n_{Mg\left(OH\right)_2}=n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\\n_{Fe\left(OH\right)_2}=n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{\downarrow}=0,3.58+0,1.90=26,4\left(g\right)\)

a. Số gam mỗi kim loại trong hỗn hợp ban đầu:

b. Thể tích dung dịch HCl 2M đã phản ứng: 0,2 lít

c. Khối lượng kết tủa thu được khi dd A tác dụng với dung dịch NaOH dư là 0,4 gam.

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