\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right);n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ Vì:\dfrac{0,4}{4}>\dfrac{0,2}{5}\Rightarrow P.dư\\ n_{P\left(dư\right)}=0,4-\dfrac{5}{4}.0,2=0,15\left(mol\right)\\ m_{P\left(Dư\right)}=0,15.31=4,65\left(g\right)\)

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

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

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Ta có: \(\dfrac{n_P}{4}=\dfrac{0,4}{4}\)

\(\dfrac{n_{O_2}}{5}=\dfrac{0,2}{5}\)

\(\Rightarrow\dfrac{n_P}{4}>\dfrac{n_{O_2}}{5}\)

Vậy phốt pho dư

\(n_{P\text{Pứ}}=\dfrac{0,4.4}{5}=0,32\left(mol\right)\)

\(n_{Pdư}=n_P-n_{PPứ}=0,4-0,32=0,08\left(mol\right)\)

Khối lượng phốt pho dư:

\(m_{Pdư}=n_{Pdư}.M_P=0,08.31=2,48g\)

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

$n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} = \dfrac{7,84}{22,4} = 0,35(mol)$
$n_P : 4 = 0,05 < n_{O_2} :5 = 0,07$ nên $O_2$ dư

$n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)$
$\Rightarrow m_{O_2\ dư} = (0,35 - 0,25).32 = 3,2(gam)$

c) $n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)$
$m_{P_2O_5} = 0,1.142 = 14,2(gam)$

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\\a, 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ V\text{ì}:\dfrac{0,35}{5}>\dfrac{0,2}{4}\Rightarrow O_2d\text{ư}\\ n_{O_2\left(d\text{ư}\right)}=0,35-\dfrac{5}{4}.0,2=0,1\left(mol\right)\\b, m_{O_2\left(d\text{ư}\right)}=0,1.32=3,2\left(g\right)\\ c,n_{P_2O_5}=\dfrac{n_P}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{r\text{ắn}}=m_{P_2O_5}=142.0,1=14,2\left(g\right)\)

nP = 6.2/31 = 0.2 (mol) 

nO2 = 6.72/22.4 = 0.3 (mol) 

4P + 5O2 -to-> 2P2O5 

0.2___0.25_____0.1

mO2 dư = ( 0.3 - 0.25) * 32 = 1.6(g) 

mP2O5 = 0.1*142 = 14.2 (g) 

Ta có: \(n_P=\dfrac{6.2}{31}=0.29mol\)

\(n_{O_2}=\dfrac{6.72}{22.4}=0.3mol\)

PTHH:

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

ta có:

\(\left\{{}\begin{matrix}\dfrac{n_{P\left(bra\right)}}{nP_{\left(pthh\right)}}=\dfrac{0.2}{4}=0.05\\\dfrac{n_{O_2\left(bra\right)}}{n_{O_2}\left(pthh\right)}=\dfrac{0.3}{5}=0.06\end{matrix}\right.\)

=> \(O_2\) dư 

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

4   ----------->2                

0.2---------->0.1=n_P2O5

=>\(m_{P_2O_5}=142.0.1=14.2\left(g\right)\)

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

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\)

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

Xét: \(\dfrac{0,2}{4}\) < \(\dfrac{0,3}{5}\)                         ( mol )

         0,2                          0,1        ( mol )

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

`PTHH: 4P + 5O_2` $\xrightarrow[]{t^o}$ `2P_2 O_5`

`n_P = [ 6,2 ] / 31 = 0,2 (mol)`

`n_[O_2] = [ 6,72 ] / [ 22,4 ] = 0,3 (mol)`

Ta có: `[ 0,2 ] / 4 < [ 0,3 ] / 5`

   `->P` hết ; `O_2` dư

Theo `PTHH` có: `n_[P_2 O_5] = 1 / 2 n_P = 1 / 2 . 0,2 = 0,1 (mol)`

         `-> m_[P_2 O_5] = 0,1 . 142 = 14,2 (g)`

\(a.4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ n_P=\dfrac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\dfrac{19,2}{32}=0,6\left(mol\right)\\ Vì:\dfrac{0,2}{4}< \dfrac{0,6}{5}\\ \Rightarrow O_2dư\\ \Rightarrow n_{O_2\left(dư\right)}=0,6-\dfrac{5}{4}.0,2=0,35\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,35.32=11,2\left(g\right)\\ b,n_{P_2O_5}=\dfrac{n_P}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\)

Tham khảo

nP = 6.2/31 = 0.2 (mol) 

nO2 = 6.72/22.4 = 0.3 (mol) 

4P + 5O2 -to-> 2P2O5 

0.2___0.25_____0.1

mO2 dư = ( 0.3 - 0.25) * 32 = 1.6(g) 

mP2O5 = 0.1*142 = 14.2 (g) 

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

nP=\(\dfrac{62}{31}\)=0,2(mol)

nO2=\(\dfrac{7,84}{22,4}\)=0,35(mol)

PTHH:4P+5O2to→2P2O5

tpứ:    0,2     0,35     

pứ:     0,2     0,25     0,1

spứ:     0     0,1         0,1

a)chất còn dư là oxi

mO2dư=0,1.32=3,2(g)

b)mP2O5=n.M=0,1.142=14,2(g)

\(a.n_P=0,2\left(mol\right);n_{O_2}=0,35\left(mol\right)\\ 4P+5O_2-^{t^o}\rightarrow2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,35}{5}\\ \Rightarrow SauphảnứngO_2dư\\ n_{O_2\left(pứ\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\\ \Rightarrow m_{P\left(dư\right)}=\left(0,35-0,25\right).32=3,2\left(g\right)\\ b.n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

a) n\(_P\) = \(\dfrac{6,2}{31}\) = 0,2(mol); n\(_{O_2}\) = \(\dfrac{7,84}{22,4}\) = 0,35(mol).

Theo phương trình hóa học:

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

4        5          2 (mol)

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

Tỉ lệ số mol: \(\dfrac{0,2}{4}< \dfrac{0,35}{5}\) \(\Rightarrow\) \(O_2\) phản ứng dư, P phản ứng hết.

\(n_{O_2}\)dư = 0,35 - \(\dfrac{0,2.5}{4}\) = 0,1(mol) \(\Rightarrow\) \(m_{O_2}\)dư = 0,1.32 = 3,2g

b) \(n_{P_2O_5}\) = 0,1.142 = 14,2g

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)

\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)

\(a) n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} =\dfrac{5,6}{22,4} = 0,25(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,05 = \dfrac{n_{O_2}}{5} = 0,05 \Rightarrow \text{Không có chất nào dư}\\ b) n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)\\ m_{P_2O_5} = 0,1.142 = 14,2(gam) \)