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

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

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) 

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

$n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} = \dfrac{7,84}{22,4} = 0,35(mol)$
$n_P : 4 = 0,05 < n_{O_2} :5 = 0,07$ 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) $n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)$
$m_{P_2O_5} = 0,1.142 = 14,2(gam)$

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

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

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

a, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)

b, \(V_{kk}=5V_{O_2}=28\left(l\right)\)

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

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

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

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

a) Theo PTHH:

nO2=54nP=54.0,2=0,25nO2=54nP=54.0,2=0,25 (mol)

Thể tích khí oxi tham gia phản ứng (đktc) là:

VO2=0,25.22,4=5,6VO2=0,25.22,4=5,6 (l)
b) 

  (mol)

Theo PTHH:

nP2O5=12nP=12.0,2=0,1nP2O5=12nP=12.0,2=0,1 (mol)

Khối lượng  thu được sau phản ứng là:

mP2O5=0,1.142=14,2mP2O5=0,1.142=14,2 (g)

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

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

_____0,2-->0,25------>0,1

=> V_O2 = 0,25.22,4 = 5,6 (l)

b) m_P2O5 = 0,1.142 = 14,2 (g)

\(n_P=\dfrac{m}{M}=0,2\left(mol\right)\)

- Ta có : \(V_{O_2}=\dfrac{V_{kk}}{5}=4,48\left(l\right)\)

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

\(4P+5O_2\rightarrow2P_2O_5\)

- Theo phương pháp đường chéo ta có :

=> Sau phản ứng O₂ phản ứng hết, P còn dư ( dư 0,04 mol )

Vậy sau phản ứng photpho không cháy hết .

b, - Chất được tạo thành là P2O5 .

Theo PTHH : \(n_{P2O5}=\dfrac{n_P}{2}=\dfrac{0,16}{2}=0,08\left(mol\right)\)

\(\Rightarrow m_{P2O5}=n.M=11,36\left(g\right)\)

\(n_{KClO_3}=\dfrac{29.4}{122.5}=0.24\left(mol\right)\)

\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)

\(0.24.....................0.36\)

KClO₃ : Kali clorat

KCl : Kali clorua

\(V_{O_2}=0.36\cdot22.4=8.064\left(l\right)\)

\(b.\)

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

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

Lập tỉ lệ : 

\(\dfrac{0.2}{4}< \dfrac{0.36}{5}\) => O₂ dư

\(n_{O_2\left(dư\right)}=0.36-0.2\cdot\dfrac{5}{4}=0.11\left(mol\right)\)

\(m_{O_2}=0.11\cdot32=3.52\left(g\right)\)

\(m_{P_2O_5}=0.1\cdot142=14.2\left(g\right)\)

Chúc em học tốt nhé !

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

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

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