`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`

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

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

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.

`#3107.101107`

1.

a.

Ta có:

\(\text{n}_{\text{KClO}_3}=\dfrac{\text{m}_{\text{KClO}_3}}{\text{M}_{\text{KClO}_3}}=\dfrac{122,5}{122,5}=1\text{ (mol)}\)

PTPỨ: \(\text{2KClO}_3\text{ }\)\(\underrightarrow{\text{ }t^0}\) \(\text{2KCl}+3\text{O}_2\) 

Ta có: `2` mol \(\text{KClO}_3\) thu được `3` mol \(\text{O}_2\)

`=>` `1` mol \(\text{KClO}_3\) thu được `1,5` mol \(\text{O}_2\)

b.

\(\text{V}_{\text{O}_2}=\text{n}_{\text{O}_2}\cdot24,79=1,5\cdot24,79=37,185\left(l\right)\)

TTĐ:

\(m_{KClO_3}=122,5\left(g\right)\)

______________

a) PTHH?

b) \(V_{O_2}=?\left(l\right)\)

                                                       Giải

                                \(n_{KClO_3}=\dfrac{m}{M}=\dfrac{122,5}{122,5}=1\left(mol\right)\)

                                    \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)

                                    1->             1        :  1,5(mol)

                              \(V_{O_2}=n.22,4=1,5.22,4=33,6\left(l\right)\)

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

Ta có: \(n_{CH_4}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)

a, \(n_{H_2O}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow m_{H_2O}=2,5.18=45\left(g\right)\)

b, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow V_{O_2}=2,5.24,79=61,975\left(l\right)\)

Mà: O₂ chiếm 1/5 thể tích không khí.

\(\Rightarrow V_{kk}=5V_{O_2}=309,875\left(l\right)\)

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

Ta có: \(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)

\(n_{O_2}=\dfrac{40}{32}=1,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,8}{4}< \dfrac{1,25}{5}\), ta được O₂ dư.

Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,4\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,4.142=56,8\left(g\right)\)

\(n_{O_2}=\dfrac{V_{O_2}}{24,79}=\dfrac{6,21}{24,79}=0,25mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

0,2     0,25              0,1      ( mol )

\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,1.142=14,2g\)

\(m_P=n_P.M_P=0,2.31=6,2g\)