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

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

PTHH:\(4P+5O_2\rightarrow2P_2O_5\)

tpứ:    0,2     0,35     

pứ:     0,2     0,25     0,1

spứ:     0     0,1         0,1

a)chất còn dư là oxi

\(m_{O_2dư}=n.M\)=0,1.32=3,2(g)

b)\(m_{P_2O_5}=n.M\)=0,1.142=14,2(g)

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

\(n_{O_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)

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

\(4........5\)

\(0.2........0.35\)

\(LTL:\dfrac{0.2}{4}< \dfrac{0.35}{5}\Rightarrow O_2dư\)

\(m_{O_2\left(dư\right)}=\left(0.35-0.25\right)\cdot32=3.2\left(g\right)\)

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

Tham khảo nha!!!

nP = 6,2/31 = 0,2 mol ; nO2 = 7,84/22,4 = 0,35 mol

a, PTHH :  4P  +  5O2 (to) -> 2P2O5 

                 0,2      0,35 mol

Ta thấy : 0,2/4 < 0,35/5 -> nO2 dư = 0,35 - 0,05*5 = 0,1 mol

-> mO2 dư = 0,1*32 = 3,2 gam

b, Theo pt : nP2O5 = 1/2*nP = 0,1 mol -> mP2O5 = 0,1*142 = 14,2 gam

Bài 7 :

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

\(n_{O2}=\dfrac{V}{22,4}=0,35\left(mol\right)\)

a, \(PTHH:4P+5O_2\rightarrow2P_2O_5\)

- Xét PTHH => Sau phản ứng O2 dư dư ( \(0,35-0,2.\dfrac{5}{4}=0,1\left(mol\right)\) )

=> \(m_{O2du}=n.M=3,2\left(g\right)\)

b, - Theo PTHH : \(n_{P2O5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)

\(\Rightarrow m_{P2O5}=n.M=14,2\left(g\right)\)

Vậy ...

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

C9:

nP = 6,2/31 = 0,2 (mol)

nO2 = 6,4/32 = 0,2 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5

LTL: 0,2/4 > 0,2/5 => P dư

nP (p/ư) = 0,2/5 . 4 = 0,16 (mol)

nP (dư) = 0,2 - 0,16 = 0,04 (mol)

nP2O5 = 0,2/5 . 2 = 0,08 (mol)

mP2O5 = 0,08 . 142 = 11,36 (g)

C10: 

Áp dụng ĐLBTKL, ta có:

mR + mO2 = mRO

=> mO2 = 21,6 - 16,8 = 4,8 (g(

=> nO2 = 4,8/32 = 0,15 (mol)

PTHH: 2R + O2 -> (t°) 2RO

nR = 0,15 . 2 = 0,3 (mol)

M(R) = 16,8/0,3 = 56 (g/mol(

=> R là Fe

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

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

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

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

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

nP= 0.1 mol

nO2= 0.15 mol

4P + 5O2 -to-> 2P2O5

4____5

0.1___0.15

Lập tỉ lệ: 0.1/4 < 0.15/5 => O2 dư

nO2 dư= 0.15 - 0.125=0.025 mol

mO2 dư= 0.8g

nP2O5= 0.05 mol

mP2O5= 7.1g

\(n_P=\dfrac{m_P}{M_P}=\dfrac{12,4}{31}=0,4mol\)

\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{17}{32}=0,53125mol\)

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

0,4 < 0,53125                     ( mol )

0,4     0,5                 0,2         ( mol )

\(n_{O_2\left(du\right)}=0,53125-0,5=0,03125mol\)

Chất được tạo thành là P2O5

\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,2.142=18,4g\)