Đáp án cần chọn là: D

2Al + 6HCl -> 2AlCl₃ + 3H₂ (1)

ZnO + 2HCl -> ZnCl₂ + H₂O (2)

a) n_H2= 13,44/22.4=0.6(mol) -> mH2=0,6.2=1,2(g)

Theo PTHH: n_Al = 2/3 n_H2 = 2/3 . 0,6= 0,4(mol) -> m_Al = 0,4 . 27=10,8(g)

-> m_ZnO = 27-10,8= 16,2(g)

b) n_ZnO = 16,2/81=0,2(mol)

Theo PTHH (2): nHCl = 2nZnO=2.0,2=0,4(mol)

Theo PTHH (1) : nHCl=2nH₂=2.0,6=1,2(mol)

-> \(\Sigma\)nHCl = 0,4+1,2=1,6(mol)

-> mHCl = 1,6.36,5= 58,4(g)

-> mdd_HCl = 58,4.100/29,2= 200(g)

c) Theo PTHH (1): nAlCl₃ = 2/3 nH₂ = 2/3 . 0,6=0,4(mol)                                         -> mAlCl3=0,4.133,5=53,4(g)

mdd sau phản ứng= m_A + mddHCl - mH2 =27+200-1,2 =225,8(g)

-> C% AlCl3 = 53,4.100%/225,8 = 20,88%

Theo PTHH (2) nZnCl2 =nZnO= 0,2(mol)-> mZnCl2=0,2.136=27,2(g)

-> C% ZnCl2= 27,2.100%/255,8=10,63%

a) 2Al + 6HCl -> 2AlCl3 + 3H2

Al2O3 + 6HCl -> 2AlCl3 + 3H2O

nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol

=>%mAl=20,93% =>%mAl2O3 = 79,07%

b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g

mddY=12,9+100-0,15.2=112,6g

mAlCl3=22,5g=>C%=19,98%

\(a.n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+24b=5,1\\1,5a+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,1}.100\approx52,941\%\\\%m_{Mg}\approx47,059\%\end{matrix}\right.\)

\(b.m_{ddH_2SO_4}=\dfrac{0,25.98.100}{9,8}=250\left(g\right)\\ m_{ddsau}=m_{Al,Mg}+m_{ddH_2SO_4}-m_{H_2}=5,1+250-0,25.2=254,6\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342}{254,6}.100\approx6,716\%\\ C\%_{ddMgSO_4}=\dfrac{0,1.120}{254,6}.100\approx4,713\%\)

a)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:       x                                                     1,5x

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:      y                                                 y

Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)

b) 

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:      0,1      0,15                  0,05                            

PTHH: Mg + H₂SO₄ → MgSO₄ + H₂

Mol:     0,1       0,1                 0,1

\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)

m_{dd sau pứ} = 5,1+250-0,15.2 = 254,8(g)

\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)

\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)