\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(2A1+2NAOH+2H_2O-2NaA10_2+H_2O\)

\(AI_2O_3=2NaOH+2NaOHA10_2+H_2O\)

\(n_{AI}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)

\(m_{AI}=27.0,4=10,8\left(gam\right);mAI_2O_3=31,2-10,8=20,4\left(gam\right)\)

Biết làm mỗi câu A

 Chỉ Có Al tác dụng với HCl sp tạo ra khí H₂

\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
   0.4           \(\leftarrow\)                           0.6  (mol)

\(Al_2O_3+HCl\rightarrow AlCl_3+H_2O\)

n_Al=\(\dfrac{0.6\cdot2}{3}=0.4\left(mol\right)\)

m_Al=0.4*27=10.8(g)

=)) \(m_{Al_2O_3}=31.2-10.8=20.4\left(g\right)\)

\(\left\{{}\begin{matrix}\%Al=\dfrac{10.8}{31.2}\cdot100=34.62\%\\\%Al_2O_3=\dfrac{20.4}{31.2}\cdot100=65.38\%\end{matrix}\right.\)

Đáp án D

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%

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol ) \(\rightarrow m_{hh}=27x+56y=5,54\left(g\right)\)  (1)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

 x                                                   1,5x    ( mol )

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

 y                                           y       ( mol )

\(n_{H_2}=1,5x+y=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)   (1)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,07\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,06.27}{5,54}.100=29,24\%\\\%m_{Fe}=100-29,24=70,76\%\end{matrix}\right.\)

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)

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

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

a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH:
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\) (2)

Theo PT (1): \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Al_2O_3}=15,6-5,4=10,2\left(g\right)\end{matrix}\right.\)

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

Theo PT (1), (2): \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}+n_{Al_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{mu\text{ố}i}=m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)

c) Theo PT (1), (2): \(n_{H_2SO_4}=n_{H_2}+3n_{Al_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(c\text{ần}.d\text{ùng}\right)}=0,6.98=58,8\left(g\right)\)

Đáp án cần chọn là: D