\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

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

PTHH :

\(4Al+3O_2\rightarrow2Al_2O_3\)

2/15    0,1       1/15

  \(\dfrac{0,1}{3}< \dfrac{0,1}{2}\)

---> Tính theo O2 

\(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\\a, 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{O_2}=\dfrac{3}{4}.0,5=0,375\left(mol\right)\\ m_{O_2}=32.0,375=12\left(g\right)\\ V_{O_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

Ta có : 

\(n_{Al_2O_3}=\dfrac{0.2\cdot2}{4}=0.1\left(mol\right)\)

\(m_{Al_2O_3}=0.1\cdot102=10.2\left(g\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.2.......\dfrac{2}{15}.....\dfrac{1}{15}\)

\(V_{O_2}=\dfrac{2}{15}\cdot22.4=2.98\left(l\right)\)

\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.46\left(g\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{25,55}{36,5}=0,7\left(mol\right)\\a. 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Vì:\dfrac{0,2}{2}< \dfrac{0,7}{6}\\ \Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,7-\dfrac{6}{2}.0,2=0,1\left(mol\right)\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ m_{H_2}=0,3.2=0,6\left(g\right)\\ m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\\ m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)

n_Al = \(\dfrac{5,4}{27}=0,2\) mol

n_O2 = \(\dfrac{6,4}{32}=0,2\) mol

Pt: 4Al + 3O₂ --to--> 2Al₂O₃

0,2 mol->0,15 mol-> 0,1 mol

Xét tỉ lệ mol giữa Al và O₂:

\(\dfrac{0,2}{4}< \dfrac{0,2}{3}\)

Vậy O₂ dư

m_Al2O3 = 0,1 . 102 = 10,2 (g)

m_O2 dư = (0,2 - 0,15) . 32 = 1,6 (g)

4Al + 3O₂ + 2Al₂O₃

_{0,2 0,15 0,1}

nAl = m/M=5,4/27=0,2

mAl₂O₃₌ n.M=0,1×(27×2+16×3)=10,2g

n_Al = 0,2 mol

nO2 = 0,2 mol

4Al + 3O₂ → 2Al₂O₃

Đặt tỉ lệ ta có

\(\dfrac{0,2}{4}\) < \(\dfrac{0,2}{3}\)

⇒ O₂ dư

⇒ m_Al2O3 = 0,1.102 = 10,2 (g)

⇒ mO2 dư = ( 0,2-0,15 ).32 = 1,6 (g)

\(PTHH:4Al+3O_2->2Al_2O_3\)

BĐ          0,4      0,27                     (mol)

PU        0,36---->0,27---->0,18         (mol) 

CL          0,04---->0------>0,18     (mol)

b)

\(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,048}{22,4}=0,27\left(mol\right)\)

\(\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\left(\dfrac{0,4}{4}>\dfrac{0,27}{3}\right)\)

=> Al dư, O2 hết (tính theo O2)

\(m_{Al}=n\cdot M=0,04\cdot27=1,08\left(g\right)\)

c)

\(m_{Al_2O_3}=n\cdot M=0,18\cdot\left(27\cdot2+16\cdot3\right)=18,36\left(g\right)\)

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

b, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{6,048}{22,4}=0,27\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,4}{4}>\dfrac{0,27}{3}\), ta được Al dư.

Theo PT: \(n_{Al\left(pư\right)}=\dfrac{4}{3}n_{O_2}=0,36\left(mol\right)\)

\(\Rightarrow n_{Al\left(dư\right)}=0,4-0,36=0,04\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,04.27=1,08\left(g\right)\)

c, Theo PT: \(n_{Al_2O_3}=\dfrac{2}{3}n_{Al}=0,18\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,18.102=18,36\left(g\right)\)