4P+5O₂-to>2P₂O₅

0,05--0,0625------0,025 mol

n _P=\(\dfrac{1,55}{31}\)=0,05 mol

=>V_O2=0,0625.22,4=1,4l

=> m_P2O5=0,025.142=3,55g

\(n_P=\dfrac{1,55}{31}=0,05\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ a,n_{O_2}=\dfrac{5}{4}.0,05=0,0625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,0625.22,4=1,4\left(l\right)\\ b,n_{P_2O_5}=\dfrac{0,05}{2}=0,025\left(mol\right)\\ \Rightarrow m_{P_2O_5}=0,025.142=3,55\left(g\right)\)

Đáng ra thấy tên T.Anh anh sẽ x24,79 nhưng đề cho đkc anh x22,4 nè.

nFe = 16,8/56 = 0,3 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 0,3 ---> 0,2

VO2 = 0,2 . 22,4 = 4,48 (l)

PTHH: 4P + 5O2 -> (t°) 2P2O5

Mol: 0,16 <--- 0,2

mP = 0,16 . 31 = 4,96 (g)

Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

_____0,4____0,5_____0,2 (mol)

a, \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

b, \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

n_Fe = 16.8/56 = 0.3 (mol) 

3Fe + 2O₂ -to-> Fe₃O₄ 

0.3......0.2...........0.1

VO2 = 0.2*22.4 = 4.48 (l) 

mFe3O4 = 0.1*232 = 23.2 (g) 

a. \(n_{KMnO_4}=\dfrac{47.4}{158}=0,3\left(mol\right)\)

PTHH : 2KMnO₄ ---t^o----> K₂MnO₄ + MnO₂ + O₂

              0,3                                                      0,15

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

b. PTHH : 4Al + 3O₂ -> 2Al₂O₃

               0,2      0,15 

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

a) PTHH: 4P + 5O₂ -> 2P₂O₅

b,c) ĐLBTKL

\(m_P+m_{O_2}=m_{P_2O_5}\\ m_{O_2}=14,2-6,2=8\left(g\right)\)

a, 4P + 5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

b, Theo ĐLBTKL, ta có:

m_P + m_{O\(_2\)} = m_{\(P_2O_5\)}

c, \(\Rightarrow m_{O_2}=14,2-6,2=8g\)

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

2Zn+O2-to>2ZnO

0,1---0,05----0,1

n Zn=\(\dfrac{6,5}{65}\)=0,1 mol

n O2=\(\dfrac{0,8}{32}\)=0,025 mol

=>VO2=0,05.22,4=1,12l

=>mZnO=0,1.81=8,1g

c)Zn dư

=>m ZnO=0,05.81=4,05g

nP = 3,1/31 = 0,1 (mol)

PTHH: 4P + 5O2 -t°-> 2P2O5

             0,1---> 0,125--->0,05

VO2 = 0,125 . 22,4 = 2,8 (l)

mP2O5 = 0,05 . 142 = 7,1 (g)

\(n_P=\dfrac{3,1}{31}=0,1mol\)

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

0,1     0,125   0,05

\(V_{O_2}=0,125\cdot22,4=2,8l\)  

\(m_{P_2O_5}=0,05\cdot142=7,1g\)

a,PTHH : \(4P+5O_2->2P_2O_5\)

b, 

Số mol O2 tham gia phản ứng :

nO2 = 5/4.n_P = 5/4 . 0,1 = 0,125 ( mol )

=> V O₂ phản ứng : 0,125 . 22,4 = 2,8 ( lít )

c,

Số mol P2O5 : n_p2o5 = 1/2.n_p = 0,05 ( mol )

=> m = 142 . 0,05 = 7,1 (g)