Đáp án : A

n_H2 = 0,25 mol; n_SO2 = 0,3 mol

2H⁺   + 2e → H₂                                  S⁺⁶ + 2e   → S⁺⁴

           0,5 <-- 0,25                                      0,6 <-- 0,3

n_Fe = 0,6 – 0,5 = 0,1 mol

=> m_Fe = 5,6g

Đặt \(n_{Fe}=a\left(mol\right)\) \(\Rightarrow n_{Al}=2a\left(mol\right)\)

Ta có: \(56a+27\cdot2a=11\) \(\Leftrightarrow a=0,1\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

Bảo toàn electron: \(3n_{Al}+2n_{Fe}=2n_{H_2}\)

\(\Rightarrow n_{H_2}=0,4\left(mol\right)\) \(\Rightarrow V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\)

Đáp án A

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

=> \(n_{H_2SO_4}=0,2\left(mol\right)\)

m_muối = m_{kim loại} + m_SO4 = 12 + 0,2.96 = 31,2 (g)

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

\(n_{Fe}=b\left(mol\right)\)

\(m=27a+56b=19.3\left(g\right)\left(1\right)\)

\(n_{H^+}=0.2\cdot2+0.2\cdot2.25\cdot2=1.3\left(mol\right)\)

\(2Al+6H^+\rightarrow2Al^{3+}+3H_2\)

\(Fe+2H^+\rightarrow Fe^{2+}+H_2\)

\(n_{H^+}=3a+2b=1.3\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.2\)

\(\%Al=\dfrac{0.3\cdot27}{19.3}\cdot100\%=41.96\%\)

\(\%Fe=58.04\%\)

\(b.\)

\(n_{H_2}=\dfrac{1}{2}n_{H^+}=0.65\left(mol\right)\)

Bảo toàn khối lượng : 

\(m_{Muối}=19.3+0.4\cdot36.5+0.45\cdot98-0.65\cdot2=76.7\left(g\right)\)

em cảm ơn ạ!

Gọi số mol Al và Fe lần lượt là a;b

$\Rightarrow 27a+56b=8,3$

Bảo toàn e ta có: $3a+3b=0,6$

Giải hệ ta được $a=b=0,1$

$\Rightarrow m_{Al}=2,7(g);m_{Fe}=5,6(g)$

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

\(m=27a+56b=8.3\left(g\right)\left(1\right)\)

\(n_{SO_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(BTe:\)

\(3a+3b=0.3\cdot2=0.6\left(2\right)\)

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

\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

\(m_{Fe}=5.6\left(g\right)\)

Trong \(20,4g\) hỗn hợp có: \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\\n_{Al}=c\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow65a+56b+27c=20,4\left(1\right)\)

\(n_{H_2}=\dfrac{10,08}{22,4}=0,45mol\)

\(BTe:2n_{Zn}+2n_{Fe}+3n_{Al}=2n_{H_2}\)

\(\Rightarrow2a+2b+3c=2\cdot0,45\left(2\right)\)

Trong \(0,2mol\) hhX có \(\left\{{}\begin{matrix}Zn:ka\left(mol\right)\\Fe:kb\left(mol\right)\\Al:kc\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow ka+kb+kc=0,2\)

\(n_{Cl_2}=\dfrac{6,16}{22,4}=0,275mol\)

\(BTe:2n_{Zn}+3n_{Fe}+3n_{Al}=2n_{Cl_2}\)

\(\Rightarrow2ka+3kb+3kc=2\cdot0,275\)

Xét thương:

 \(\dfrac{ka+kb+kc}{2ka+3kb+3kc}=\dfrac{0,2}{2\cdot0,275}\Rightarrow\dfrac{a+b+c}{2a+3b+3c}=\dfrac{4}{11}\)

\(\Rightarrow3a-b-c=0\left(3\right)\)

Từ (1), (2), (3)\(\Rightarrow\left\{{}\begin{matrix}a=0,1mol\\b=0,2mol\\c=0,1mol\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Zn}=6,5g\\m_{Fe}=11,2g\\m_{Al}=2,7g\end{matrix}\right.\)

Gọi x, y lần lượt là số mol Al, Fe

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}27x+56y=0,83\\1,5x+y=0,025\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,27\left(g\right)\\m_{Fe}=0,56\left(g\right)\end{matrix}\right.\)