a) 4Al + 3O₂ --t^o--> 2Al₂O₃

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

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

_____0,2->0,15

=> V_O2(đktc) = 0,15.22,4 = 3,36 (l)

=> m_O2 = 0,15.32 = 4,8(g)

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

cảm ơn bạn rất nhiều 

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: \(m_{Al}+m_{O_2}=m_{Al_2O_3}\)

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

thank bạn

a. 4Al + 3O2 -> 2Al2O3

     0.3    0.225     0.15

b.\(n_{Al}=\dfrac{8.1}{27}=0.3mol\)

\(V_{O_2}=0.225\times22.4=5.04l\)

c.\(m_{Al_2O_3}=0.15\times102=15.3g\)

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

a)

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

b)

Bảo toàn khối lượng: 

\(m_{Al} + m_{O_2} = m_{Al_2O_3}\\ \Rightarrow m_{O_2} = 10,2 - 5,4 = 4,8(gam)\)

d)

Theo PTHH : \(n_{Al_2O_3} = 0,5n_{Al} = 0,1(mol)\)

Sao có 0,5n_al     vậy ạ?

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

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

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)