\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{Cl_2}=\dfrac{V_{Cl_2\left(ĐKTC\right)}}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+3Cl_2\rightarrow2AlCl_3\\ \dfrac{n_{Al}}{n_{Cl_2}}=\dfrac{1}{2}< \dfrac{2}{3}\Rightarrow Cl_2\text{ dư, bài toán tính theo }Al\\ \Rightarrow m_{AlCl_3}=n_{AlCl_3}\cdot M_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ a,m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\\ n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ c,C_{MddHCl}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

nAl = 2,7/27 = 0,1 (mol)

PTHH: 2Al + 3Cl2 -> (t°) 2AlCl3

Mol: 0,1 ---> 0,15 ---> 0,1

VCl2 = 0,15 . 22,4 = 3,36 (l)

mAlCl3 = 0,1 . 133,5 = 13,35 (g)

a/

2Al+3Cl2 --(t^o)--> 2AlCl3

\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)

=> \(nCl_2=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)

\(VCl_2=0,15.22,4=3,36\left(lít\right)\)

c, \(nAlCl_3=nAl=0,1\left(mol\right)\)

\(mAlCl_3=0,1.133,5=13,35\left(g\right)\)

a) nHCl=14,6/36,5=0,4(mol); nAl=2,7/27=0,1(mol)

PTHH: 2 Al + 6 HCl -> 2 AlCl3 +3 H2

Ta có: 0,1/2 < 0,4/6

=>Al hết, HCl dư, tính theo nAl

nHCl(dư)= 0,4- 0,1.6/2=0,1(mol)

=>mHCl(dư)=0,1.36,5=3,65(g)

b) nAlCl3=nAl=0,1(mol)

=> mAlCl3=133,5.0,1=13,35(g)

2Al+6HCl----->2AlCl3+3H2

nAl=2,7/27=0,1 mol

cứ  2mol Al------> 2 mol AlCl3 

              0,1mol ----->0,1 mol

mAlCl3=0,1.133,5=13.35g

H%=80%------->mAlCl3 thực tế thu được =13,35.80/100=10,68g

n Al = 2,7/27 =0,1 mol

2Al + 6HCl ---> 2AlCl3 + 3H2

0,1 --->                0,1 

ADCT : H% =80% ----> m AlCl3 tạo thành = (m AlCl3 thực tế  . H% ) / 100

hay   m AlCl3 = ((0,1.133,5) .80 ) / 100 =10,68 % 

a)

Ta có : \(n_{Al} = \dfrac{2,7}{27} = 0,1(mol)\)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2\)

Theo PTHH : 

\(n_{AlCl_3} = n_{Al} = 0,1(mol)\\ \Rightarrow m_{AlCl_3} = 0,1.133,5 = 13,35(gam)\)

b)

\(n_{H_2} = 1,5n_{Al} = 0,15(mol)\\ \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)\)

PTHH: \(2Al+6HCl\)→\(2AlCl_3+6H_2\)

+\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

+\(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

+\(m_{AlCl_3}=0,1.133,5=13,35\left(gam\right)\)

+\(n_{H_2}=3n_{Al}=0,3\left(mol\right)\)

+\(V_{H_2}=0,3.22,4=6,72\left(mol\right)\)

n_Al  = \(\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 6HCl -> 2AlCl₃ + 3H₂ ↑

PT : 2                6           2            3     (mol)

ĐB: 0,1              0,3         0,1        0,15   (mol)

a) m_AlCl3 = n.M = 0,1.133,5 = 13,35(g)

b) V_H2 = n.22,4 = 0,15.22,4 = 3,36(l)

a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

           0,1--->0,3------>0,1---->0,15

=> m_HCl = (0,4 - 0,3).36,5 = 3,65 (g)

b) V_H2 = 0,15.22,4 = 3,36 (l)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)

\(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{14,6}{36,5}=0,4mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1  <   0,4                                 ( mol )

0,1        0,3                         0,15 ( mol )

a. Chất còn dư là HCl

\(m_{HCl}=n_{HCl}.M_{HCl}=\left(0,4-0,3\right).36,5=3,65g\)

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36l\)

\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

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

0,1         0,15                                0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{H_2SO_4}=0,15.98=14,7g\)