a) 2Al + 3Cl₂ --t^o--> 2AlCl₃

b) \(n_{AlCl_3}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\)

PTHH: 2Al + 3Cl₂ --t^o--> 2AlCl₃

            0,1<-0,15<---------0,1

=> V_Cl2 = 0,15.22,4 = 3,36(l)

c) m_Al = 0,1.27 = 2,7(g)

\(a.PTHH:2Al+3Cl_2\overset{t^o}{--->}2AlCl_3\)

b. Ta có: \(n_{AlCl_3}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}.n_{AlCl_3}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,15.22,4=3,36\left(lít\right)\)

c. Theo PT: \(n_{Al}=n_{AlCl_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)

a) $2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
b) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
Theo PTHH : $n_{Cl_2} = \dfrac{3}{2}n_{Al} =  0,3(mol)$
$\Rightarrow V_{Cl_2} = 0,3.24,79 = 7,437(lít)$

c) $n_{AlCl_3} = n_{Al} = 0,2(mol)$
$\Rightarrow m_{AlCl_3} = 0,2.133,5 = 26,7(gam)$

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

PTHH: 2Al + 3Cl₂ --t^o--> 2AlCl₃

_____0,1-->0,15

=> V_Cl2 = 0,15.22,4 = 3,36(l)

\(n_{Al}=\frac{1,08}{27}=0,04mol\)

\(2Al+3Cl_2\rightarrow^{t^o}2AlCl_3\)

a) \(n_{Cl_2}=\frac{3}{2}.0,04=0,06mol\)

\(V_{Cl_2}=0,06.22,4=1,344l\)

b) Cách 1: \(m_{AlCl_3}=m_{Al}+m_{Cl_2}=1,08+71.0,06=5,34g\)

Cách 2: \(n_{AlCl_3}=n_{Al}=0,04mol\)

\(m_{AlCl_3}=0,04.133,5=5,34g\)

\(n_{Al}=\dfrac{1,08}{27}=0,04\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ a,n_{Cl_2}=\dfrac{3}{2}.0,04=0,06\left(mol\right)\\ V_{Cl_2\left(\text{Đ}KTC\right)}=0,06.22,4=1,344\left(l\right)\\ b,C1:m_{AlCl_3}=m_{Al}+m_{Cl_2}=1,08+71.0,06=5,34\left(g\right)\\ C2:n_{AlCl_3}=n_{Al}=0,04\left(mol\right)\\ m_{AlCl_3}=0,04.133,5=5,34\left(g\right)\)

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

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

\(0.1........0.3............0.1......0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(m_{AlCl_3}=0.1\cdot133.5=13.35\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2-^{t^o}>2Al_2O_3\)

tỉ lệ:            4   :    3     :     2

n(mol)     0,2---->0,15---->0,1

\(m_{Al_2O_3}=n\cdot M=0,1\cdot\left(27\cdot2+16\cdot3\right)=10,2\left(g\right)\\ V_{O_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\)

câu d đề có thiếu ko ạ?

Tính theo Phương trình hóa học ấy :D 

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

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

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

_____0,2<---0,6<-----0,2<-----0,3

=> m_Al = 0,2.27 = 5,4 (g)

c) m_HCl = 0,6.36,5 = 21,9(g)

d) m_AlCl3 = 0,2.133,5 = 26,7 (g)

N_{h2 = 6,72 : 22,4 = 0,3(mol)}

    2Al + 6HCl -> 2AlCl₃ + 3H₂

Pt :   2        6           2             3(mol)       

Đb : 0,2           0,6       0,2         0,3   (mol)              

Theo PTHH 3mol H₂ = 2 mol Al

                    0,3mol Al = 0,2 mol H₂

M_Al = n*M = 0,3 * 27 = 8,1 ( g)

MHCl = n*M= 0,6 * 36,5 = 21,9 (g)

M_AlCl3= n*M= 0,2*133,5=26,7(g)

Nếu đúng thì tích cho mik cái nha :D