\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{Fe}=0,05\left(mol\right)\\ \Rightarrow m_{Fe}=0,05\cdot56=2,8\left(g\right)\\ \Rightarrow\%_{Fe}=\dfrac{2,8}{6}\cdot100\%\approx46,67\%\\ \Rightarrow\%_{Cu}\approx100\%-46,67\%=53,33\%\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

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

 0,05                              0,05

\(m_{Fe}=0,05\cdot56=2,8g\)

\(\%m_{Fe}=\dfrac{2,8}{6}\cdot100\%=46,67\%\)

\(Cu+2HCl\) : Không phản ứng

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

Pt 0,05                                       0,05   (mol)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(m_{Fe}=n_{Fe}\cdot M_{Fe}=0,05\cdot56=2,8\left(g\right)\)

\(\%m_{Fe}=\dfrac{m_{Fe}\cdot100\%}{m_{HH}}=\dfrac{2,8\cdot100}{6}\%=\dfrac{140}{3}\%\approx47\%\)

Chọn đáp án B

Chọn đáp án B

Giải thích: Đáp án B

Số mol H₂ là 0,05 mol

PTHH: Fe + 2HCl → FeCl₂ +H₂

Suy ra n_Fe = 0,05 mol → m_Fe = 2,8 → %Fe = 56%

\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 16,8 - 6,4 = 10,4(1)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)\)

Từ (1)(2) suy ra: a = 0,1 ; b = 0,2

Vậy :

\(\%m_{Fe} = \dfrac{0,1.56}{16,8}.100\% = 33,33\%\\ \%m_{Mg} = \dfrac{0,2.24}{16,8}.100\% = 28,57\%\\ \%m_{Cu} = 100\% - 33,33\% - 28,57\% = 38,1\%\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ a,\%m_{Fe}=\dfrac{0,02.56}{4,36}.100\approx25,688\%\\ \Rightarrow\%m_{Ag}\approx74,312\%\\ b,Ta.thấy:2,18=\dfrac{1}{2}.4,36\\ \Rightarrow m_{hh\left(câuB\right)}=\dfrac{1}{2}.m_{hh\left(câuA\right)}\\ n_{Fe}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ n_{Ag}=\dfrac{2,18-0,01.56}{108}=0,015\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ 2Ag+Cl_2\rightarrow\left(t^o\right)2AgCl\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+\dfrac{1}{2}.n_{Ag}=\dfrac{3}{2}.0,01+\dfrac{1}{2}.0,015=0,0225\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,0225.22,4=0,504\left(l\right)\)

\(1)n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\)

Từ giả thiết và theo PT: 

\(\begin{cases} 24n_{Mg}+56n_{Fe}=5,2\\ n_{Mg}+n_{Fe}=0,15 \end{cases}\\ \Rightarrow n_{Mg}=0,1(mol);n_{Fe}=0,05(mol)\)

\(\Rightarrow \begin{cases} \%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%=46,15\%\\ \%m_{Fe}=100-46,15=53,85\% \end{cases}\\ 2)\Sigma n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{1}=0,3(l)=300(ml)\)