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

b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)

c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)

Giúp mik vs T-T

a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

           0,2-->0,25------->0,1

=> m_P2O5 = 0,1.142 = 14,2(g)

c) V_O2 = 0,25.22,4 = 5,6(l)

$\rm n_P=\dfrac{6,2}{31}=0,2(mol)$

$\rm a)PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5$

$\rm b)$ Theo PT: $\rm n_{O_2}=1,25n_P=0,25(mol)$

$\rm V_{O_2}=0,25.22,4=5,6(lít)$

a, Số mol P là 

n = m/M = 6,2/31 = 0,2 (mol) 

a, PTHH : 4P + 5O₂ -t⁰> 2P₂O₅

4        5           2

0,2 mol -> 0,25 mol 0,1 mol 

b, Thể tích O₂ theo đktc là : 

V = n . 22,4 = 0,25 . 22,4 = 5,6 (l) 

PTHH :   4P    +    5O₂    (t^o)--->   2P₂O₅  (1)

n_P = \(\dfrac{m}{M}=\dfrac{6.2}{31}=0.2\left(mol\right)\)

Từ (1) => n_O2  =  \(\dfrac{5}{4}\)n_P  = 0.25 (mol)

=> V_O2  = n. 22,4 = 5.6 (l)

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PTHH : 4P + 5O₂ -> 2P₂O₅

=> \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)

=> \(m_{P_2O_5}=0,1.142=14,2\left(g\right)\) 

Theo ĐLBTKL

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

=> \(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\\ V_{O_2}=0,25.22,4=5,6\left(l\right)\)

3,584l

\(n_P=\dfrac{6.2}{31}=0.2\left(mol\right)\)

\(4P+5O_2\underrightarrow{^{^{t^0}}}2P_2O_5\)

\(0.2........0.25\)

\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(n_P=\dfrac{m}{M}=\dfrac{31}{31}=1\left(mol\right)\\ n_{O_2\left(dktc\right)}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

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

ti lệ:        4     :     5      :      2

n(mol)       1        0,5      

n(mol p/ư): 0,4<--0,5------>0,2 

\(\dfrac{n_P}{4}>\dfrac{n_{O_2}}{5}\left(\dfrac{1}{4}>\dfrac{0,5}{5}\right)\)

`=>` `O_2` hết, `P` dư, tính theo`O_2`

\(n_{P\left(dư\right)}=1-0,4=0,6\left(mol\right)\)

\(m_{P\left(dư\right)}=n\cdot M=0,6\cdot31=18,6\left(g\right)\\ m_{P_2O_5}=n\cdot M=0,2\cdot\left(31\cdot2+16\cdot5\right)=28,4\left(g\right)\)

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

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

\(n_{O_2}=\dfrac{5}{4}n_P=1\left(mol\right)\) \(\Rightarrow V_{O_2}=1.22,4=2,24\left(l\right)\)

b, \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,4\left(mol\right)\Rightarrow m_{P_2O_5}=0,4.142=56,8\left(g\right)\)

c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4}=2n_{O_2}=2\left(mol\right)\Rightarrow m_{KMnO_4}=2.158=316\left(g\right)\)

\(n_P=\dfrac{m}{M}=\dfrac{24,8}{31}=0,8\left(mol\right)\) 

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

                4        5          2

               0,8      1         0,4

\(a.V_{O_2}=n.24,79=1.24,79=24,79\left(l\right)\\ b.m_{P_2O_5}=n.M=0,4.\left(31.2+16.5\right)=56,8\left(g\right)\)

\(c.PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2\downarrow+O_2\uparrow\) 

                       2                 1                   1            1

                      0,8              0,4               0,4           0,4

\(m_{KMnO_4}=n.M=0,8.\left(39+55+16.4\right)=126,4\left(g\right).\)

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

Ta có: \(n_P=\dfrac{9,3}{31}=0,3\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,375\left(mol\right)\\n_{P_2O_5}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,375\cdot22,4=8,4\left(g\right)\\m_{P_2O_5}=0,15\cdot142=21,3\left(g\right)\end{matrix}\right.\)