Phương trình hóa học: 2 A l + 3 C l 2 → 2 A l C l 3

\(PTHH:2Al+3Cl_2\) → \(2AlCl_3\)

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

Theo PTHH

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

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

m C l = m A l C l 3 - m A l = 6,675 - 1,35 = 5,325 g

n C l = 3 . n A l  ⇒ số nguyên tử Cl gấp 3 số nguyên tử Al. Công thức hóa học đơn giản của nhôm clorua: A l C l 3

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

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

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

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

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

\(a,4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ \text{Số nguyên tử Al : Số phân tử }O_2=4:3\\ b,BTKL:m_{Al}+m_{O_2}=m_{Al_2O_3}\\ \Rightarrow m_{O_2}=20,4-10,8=9,6(g)\\ c,n_{O_2}=\dfrac{9,6}{32}=0,3(mol)\\ \Rightarrow V_{O_2}=0,3.22,4=6,72(l)\\ \Rightarrow V_{kk}=6,72.5=33,6(l)\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

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

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

nH₂=6.72/22,4=0,3(mol)
PTHH: 2Al +   6HCl --> 2AlCl₃ + 3H₂
bài ra:  0,2 <-- 0,6  <--   0,2   <-- 0,3  /mol
a) mHCl = 0,6.36,5=21,9(g)
b) mAl = 0.2.24 = 4,8(g)