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, \(\Rightarrow m_{O_2}=14,2-6,2=8g\)

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

a.\(n_P=\dfrac{1,55}{31}=0,05\left(mol\right)\)

PTHH : 4P + 5O₂ -> 2P₂O₅

           0,05   0,0625    0,025

\(V_{O_2}=0,0625.22,4=1,4\left(l\right)\)

b. \(m_{P_2O_5}=0,025.142=3,55\left(g\right)\)

\(a,PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(2:1:2\left(mol\right)\)

\(0,05:0,025:0,05\left(mol\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

\(b,V_{O_2}=n.22,4=0,025.22,4=0,56\left(l\right)\)

\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

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

     0,02-->0,015-->0,01

a) \(m_{Al2O3}=0,01.102=1,02\left(g\right)\)

b) \(V_{O2\left(dktc\right)}=0,015.24,79=0,37185\left(l\right)\)

`#\text{id07 - 25th8}.`

a.

Số mol của Al có trong pứ là:

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{0,54}{27}=0,02\left(\text{mol}\right)\)

PTHH: \(\text{4Al + 3O}_2\text{ }\text{ }\underrightarrow{t^0}\text{ }\text{ 2Al}_2\text{O}_3\) 

Theo pt:   4     :     2  (mol)

`=> 0,02`  :  `0,01` (mol)

`=>` n của Al₂O₃ thu được sau pứ là `0,01` mol

m của Al₂O₃ thu được sau pứ là:

\(m_{Al_2O_3}=n_{Al_2O_3}\cdot M_{Al_2O_3}=0,01\cdot\left(27\cdot2+16\cdot3\right)=0,01\cdot102=1,02\left(g\right)\)

b.

Theo pt: 4   :   3   (mol)

`=>` 0,02  :  0,015  (mol)

`=>` n của O₂ có trong pứ là `0,015` mol

V của khí O₂ tham gia pứ ở đkc là:

\(V_{O_2}=n_{O_2}\cdot24,79=0,015\cdot24,79=0,37185\left(l\right).\)

Vậy: 

a. m Al₂O₃ `=` `1,02` g

b. V O₂ `= 0,37185` l.

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

4Fe+3O2-to>2Fe2O3

0,8-----0,6------0,4 mol

n Fe=\(\dfrac{44,8}{56}\)=0,8 mol

=>VO2=0,6.22,4=13,44l

=>m Fe2O3=0,4.160=64g

d) 2KMnO4-to>K2MnO4+MnO2+O2

        1,2---------------------------------0,6

=>m KMnO4=1,2.158=189,6g

`2KClO_3->2KCl+3O_2`(to)

0,04-----------0,02-----0,06

`n_(KClO_3)=(4,9)/(122,5)=0,04mol`

=>`V_(O_2)=0,06.24,79=1,4847l`

c)

`4P+5O_2->2P_2O_5`(to)

0,048----0,06 mol

`=>m_P=0,048.31=1,488g`

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

a) \(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\)

Theo PTHH: \(n_{P_2O_5}=\dfrac{0,04\cdot2}{4}=0,02\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=n_{P_2O_5}\cdot M_{P_2O_5}=0,02\cdot142=2,84\left(g\right)\)

b) Theo PTHH: \(n_{O_2}=\dfrac{0,04\cdot5}{4}=0,05\left(mol\right)\) 

\(\Rightarrow V_{O_2\left(dkc\right)}=n_{O_2}\cdot24,79=0,05\cdot24,79=1,2395\left(l\right)\)