Question

Answer 1

Bài 7 : 

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

        4         3           2

       0,2     0,15      0,1

a) \(n_{O2}=\dfrac{0,2.3}{4}=0,15\left(mol\right)\)

\(V_{O2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{Al2O3}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)

⇒ \(m_{Al2O3}=0,1.102=10,2\left(g\right)\)

 Chúc bạn học tốt

Answer 2

câu 7 ,8 nha

Answer 3

4Al+3O₂-to>2Al₂O₃

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

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

=>V_O2=0,15.22,4=3,36l

=>m _Al2O3=0,1.102=10,2g

bài 8

4Al+3O₂-to>2Al₂O₃

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

n _O2=\(\dfrac{3,36}{22,4}\)=0,15 mol

=>m _Al=0,2.27=5,4g

=>m _Al2O3=0,1.102=10,2g

 

 

Answer 4

Bài 8 : 

\(n_{O2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

        4        3            2

       0,2    0,15       0,1

a) \(n_{Al}=\dfrac{0,15.4}{3}=0,2\left(mol\right)\)

⇒ \(m_{Al}=0,2.27=5,4\left(g\right)\)

b) \(n_{Al2O3}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)

⇒ \(m_{Al2O3}=0,1.102=10,2\left(g\right)\)

 Chúc bạn học tốt