Đáp án C.

Kim loại không phản ứng với H₂SO₄ loãng là Cu.

Gọi n_Cu = x, n_Mg = y, n_Al = z

Ta có:

64x + 24y + 27z = 33,2 (1)

Bảo toàn e:

2n_Mg + 3n_Al = 2n_H2  

=> 2y + 3z = 2.1 (2)

2n_Cu = 2n_SO2  =>  x = 0.2 (mol) (3)

Từ 1, 2, 3 => x = 0,2; y = z = 0,4 (mol)

m_Cu = 0,2.64 = 12,8 (g)

m_Mg = 0,4.24 = 9,6 (g)

m_Al = 10,8 (g)

2Al + 2H₂O + 2NaOH→ 3H₂ + 2NaAlO₂

0,2mol 0,3mol

mAl=0,2.27=5,4g

2Al + 6HCl→ 2AlCl₃+ 3H₂

0,2mol 0,3mol

Fe + 2HCl→ FeCl₂+ H₂

0,15mol 0,45-0,3 mol

mFe=0,15.56=8,4g

mCu=32,8-(6,4+8,4)=18g

%mFe=\(\frac{8,4}{32,8}.100=25,6\%\)

%mCu=\(\frac{18}{32,8}.100=54,8\%\)

%mAl=19,6%

Đáp án là D. 38,55%.

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{Al}=0,2(mol)\Rightarrow m_{Al}=0,2.27=5,4(g)\\ \Rightarrow m_{Cu}=10-5,4=4,6(g)\\ \Rightarrow \%_{Al}=\dfrac{5,4}{10}.100\%=54\%\\ \Rightarrow \%_{Cu}=100\%-54\%=46\%\\ n_{H_2SO_4}=0,3(mol)\Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,3.98}{20\%}=147(g)\)

a. PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

Cu + H₂SO₄ ---x--->

b. Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.\dfrac{6,72}{22,4}=0,2\left(mol\right)\)

=> \(m_{Al}=0,2.27=5,4\left(g\right)\)

=> \(m_{Cu}=10-5,4=4,6\left(g\right)\)

c. \(\%_{m_{Al}}=\dfrac{5,4}{10}.100\%=54\%\)

\(\%_{m_{Cu}}=100\%-54\%=46\%\)

d. Theo PT: \(n_{H_2SO_4}=n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)

\(Al\rightarrow Al^{3+}+3e\\ Cu\rightarrow Cu^{2+}+2e\\ N^{+5}+3e\rightarrow N^{+2}\\ Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Cu}=y\left(mol\right)\end{matrix}\right.\\ n_{NO}=\dfrac{31,36}{22,4}=1,4\left(mol\right)\\ BTe:3x+2y=1,4.3\\ Mặtkhác:27x+64y=65,4\\ \Rightarrow\left\{{}\begin{matrix}x=1\\y=0,6\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{0,6.64}{65,4}.100=58,72\%\\ \Rightarrow ChọnC\)

Đáp án  B