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

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

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: \(n_{AlCl_3}=n_{Al}=0,04mol\)

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

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

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

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

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

\(PTHH:2Al+3Cl_2\) → \(2AlCl_3\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{1,35}{27}=0,05\left(mol\right)\)

Theo PTHH

⇒ \(n_{Cl_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\)

⇒ \(V_{Cl_2\left(đktc\right)}=n.22,4=0,075.22,4=1,68\left(l\right)\)

Đáp án B

2Al + 3 Cl 2   →   2 AlCl 3

Dễ thấy m chất rắn tăng = m Cl 2  = 7,1g

=> n_Cl2 = 0,1 mol

n_Al = 2/3  n Cl 2  = 1/15 mol

=> m_Al = n_Al. M_Al = 1/15. 27 = 1,8g

a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,225\left(mol\right)\Rightarrow V_{O_2}=0,225.22,4=5,04\left(l\right)\)

c, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)