a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_{P_2O_5}=\dfrac{34,08}{142}=0,24\left(mol\right)\)

4P + 5O₂ --t^o--> 2P₂O₅

0,48<-0,6<------0,24

=> m_O2 = 0,6.32 = 19,2 (g)

c)

C1: m_P = 0,48.31 = 14,88(g)

C2:

Theo ĐLBTKL: m_P + m_O2 = m_P2O5

=> m_P = 34,08-19,2 = 14,88(g)

d)

V_O2 = 0,6.22,4 = 13,44 (l)

=> V_kk = 13,44 :20% = 67,2 (l)

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

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

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

a) 4P + 5O₂ --t^o--> 2P₂O₅

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

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

         0,8--------------->0,4

=> m_P2O5 = 0,4.142 = 56,8 (g)

c)

PTHH: P₂O₅ + 3H₂O --> 2H₃PO₄

               0,4--------------->0,8

=> m_H3PO4 = 0,8.98 = 78,4 (g)

mong mọi người giúp nhanh ạ

Cảm ơn bạn nha

4P2+5)2 ---> 2P2O5

Lần lượt tính mol các chất theo tỉ lệ là đc

d) P2O5 + 3H2O ---> 2H3PO4

a)

\(n_{P_2O_5} = \dfrac{42,6}{142} = 0,3(mol)\\ \)

4P      +      5O₂ \(\xrightarrow{t^o}\)         2P₂O₅

0,6.............0,75.................0,3..........(mol)

m_P = 0,6.31 = 18,6(gam)

b)

2KClO₃  \(\xrightarrow{t^o}\)   2KCl     +    3O₂

0,5....................................0,75.....(mol)

\(m_{KClO_3} = 0,5.122,5 = 61,25(gam)\)

c)

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)

\(n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)\\ \dfrac{n_{Fe}}{3} = 0,1 < \dfrac{n_{O_2}}{2} = 0,375\)

nên hiệu suất tính theo số mol Fe.

\(n_{Fe\ pư} = 0,3.90\% = 0,27(mol)\\ n_{Fe_3O_4} =\dfrac{1}{3}n_{Fe\ pư} = 0,09(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,09.232 = 20,88(gam)\)

4P+5O₂-to>2P₂O₅

0,4--0,5-----------0,2 mol

P₂O₅+3H₂O->2H₃PO₄

0,2------------------0,4

n O₂=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>m _P=0,4.31=12,4g

=>m H_3PO4=0,4.98=39,2g

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

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

4P + 5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

\(\dfrac{0,1}{4}< \dfrac{0,2}{5}\) => O₂ dư, Photpho đủ

\(n_{O_2}=0,2-0,04=0,16\left(mol\right)\)

\(m_{P_2O_5}=\) 0,05 . 142 = 7,1 ( g )

a)

\(n_{O_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_P = \dfrac{4}{5}n_{O_2} = 0,4(mol)\\ \Rightarrow m_P = 0,4.31 = 12,4(gam)\)

b)

\(n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)\)

c)

\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5.2 = 1(mol)\\ \Rightarrow m_{KMnO_4} = 1.158 = 158(gam)\)

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

\(b.n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(\Rightarrow n_P=\dfrac{0,5}{5}.4=0,4\left(mol\right)\\ \Rightarrow m_P=0,4.31=12,4\left(g\right)\)

\(m_{O_2}=0,5.32=16\left(g\right)\\ \Rightarrow m_P+m_{O_2}=m_{P_2O_5}\\ m_{P_2O_5}=24,8+16=40,8\left(g\right)\)

a) \(PTHH:4P+5O_2\) → \(2P_2O_5\)

b) \(n_{O_2}=\dfrac{V_{O_2\left(đktc\right)}}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PTHH:

⇒ \(n_P=\dfrac{4}{5}.n_{O_2}=\dfrac{4}{5}.0,5=0,4\left(mol\right)\)

⇒ \(m_P=n.M=0,4.31=12,4\left(g\right)\)

c) Theo định luật bảo toàn khối lượng

⇒ \(m_P+m_{O_2}=m_{P_2O_5}\)

⇒ \(m_{P_2O_5}=?\)

a) 4P + 5O2 -> 2P2O5

0,4mol<-0,5mol->0,2mol(1)

0,8mol              ->0,4mol(2)

nO2=11,2/22,4=0,5mol

nP=24,8/31=0,8mol

b)mP=0,4x31=12,4g

c)mP2O5=0,4x142=56,8