\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:4Al+3O_2\rightarrow^{t^o}2Al_2O_3\\ \Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}V=V_{O_2}=0,3\cdot22,4=6,72\left(l\right)\\a=m_{Al_2O_3}=0,2\cdot102=20,4\left(g\right)\end{matrix}\right.\)

nAl2O3= 10,2/102= 0,1(mol)

a) PTHH: Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O

0,1_______0,6_______0,2_________0,3(mol)

mHCl=0,6.36,5= 21,9(g)

=>mddHCl= (21,9.100)/7,3=300(g)

b) mddsau= mAl2O3 + mddHCl= 10,2+300=310,2(g)

c) mAlCl3= 133,5.0,2=26,7(g)

=>C%ddAlCl3= (26,7/310,2).100=8,607%

4Al + 3O₂ --t^o--> 2Al₂O₃

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

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

___________0,15<------0,1

=> m_O2 = 0,15.32 = 4,8(g)

Bảo toàn KL: \(m_{Al}+m_{O_2}=m_{Al_2O_3}\)

\(\Rightarrow m_{O_2}=10,2-9=1,2(g)\)

\(\left(1\right).4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\\ \left(2\right).m_{Al}+m_{O_2}=m_{Al_2O_3}\\ \left(3\right).m_{O_2}=m_{Al_2O_3}-m_{Al}=10,2-5,4=4,8\left(g\right)\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

0,4      0,3                 0,2     ( mol )

\(m_{Al}=n_{Al}.M_{Al}=0,4.27=10,8g\)

\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,2.102=20,4g\)

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

\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\Rightarrow nAl=\dfrac{4}{3}.0,3=0,4\left(mol\right)\)

\(mAl=0,4.27=10,8\left(g\right)\)

\(nAl_2O_3=\)\(\dfrac{2}{3}.0,3=0,2\left(mol\right)\)

\(mAl_2O_3=0,2.102=20,4\left(g\right)\)

a/ PTHH

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

b/

Ta có: \(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

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

4                   2

x                    0.1

\(=>x=\dfrac{0.1\cdot4}{2}=0.2=n_{Al}\)  

\(=>m_{Al}=0.2\cdot27=5.4\left(g\right)\)

a) Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

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

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

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

=> m = 0,1.342 = 34,2 (g)

c) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

Al2O3+3H2SO4->Al2(SO4)3+3H2O

0,1------0,3------0,1-------------------0,3

n Al2O3=0,1 mol

m Al2(SO4)3=0,1.342=34,2g

C%=\(\dfrac{0,3.98}{120}100=24,5\%\)

\(a.Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2 O\\ b.n_{Al_2O_3}=0,1\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\ c.n_{H_2SO_4}=3n_{Al_2O_3}=0,3\left(mol\right)\\ C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100=24,5\%\)

a) \(4Al+3O_2\rightarrow2Al_2O_3\)

b) \(n_{O_2}=\dfrac{19,2}{32}=0,6\left(mol\right)\)

Theo pthh: \(n_{Al}=\dfrac{4}{3}n_{O_2}=\dfrac{4}{3}.0,6=0,8\)

-> \(a=0,8.27=21,6\)

->\(b=19,2+21,6=40,8\)

$n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

0,2         0,15            0,1                (mol)

$V_{O_2} = 0,15.22,4 = 3,36(lít)$
$m_{Al_2O_3} =0,1.102 = 10,2(gam)$

a) 4Al + 3O₂ --t^o--> 2Al₂O₃

Tỉ lệ Al : O₂ = 4:3

b) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)=>m_{O_2}=0,15.32=4,8\left(g\right)\)

Áp dụng ĐLBTKL:

\(m_{Al}+m_{O_2}=m_{Al_2O_3}\)

=> \(m_{Al}=10,2-4,8=5,4\left(g\right)\)

c) \(a=\dfrac{5,4.100}{90}=6\left(g\right)\)