a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,05<-----------0,05---->0,075

=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)

=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)

b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          0,05->0,0375

           2Cu + O₂ --t^o--> 2CuO 

            0,2-->0,1

=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)

Đáp án B

nH₂ = 0,4 => nHCl = 0,8 = nCl

BTKL => mA = m_muối – mCl = 39,6 – 0,8.35,5 = 11,2g

\(a,n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\\ Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}95a+133,5b=10,475\\a+1,5b=0,115\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\\ \%m_{Mg}=\dfrac{0,04.24}{0,04.24+0,05.27}.100\approx41,558\%\Rightarrow\%m_{Al}\approx58,442\%\\ b,n_{HCl}=2.n_{H_2}=2.0,115=0,23\left(mol\right)\\ \Rightarrow x=C\%_{ddHCl}=\dfrac{0,23.36,5}{100}.100=8,395\%\)

Đáp án B

BTNT "H" ta có: nHCl = 2nH2 mol.

Đặt nH2 = x mol => nHCl = 2x mol

BTKL: m KL + mHCl = m muối + mH2

=> 5 + 2x.36,5 = 5,71 + 2x => x = 0,01

=> V_H2 = 0,01.22,4 = 0,224 lit