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

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

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

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

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

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

Cảm ơn bạn nha

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

Lần lượt tính mol các chất theo tỉ lệ là đc

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

a, \(n_{P_2O_5}=\dfrac{21,3}{142}=0,15\left(mol\right)\)

PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

Theo PT: \(n_{H_3PO_4}=2n_{P_2O_5}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_3PO_4}=0,3.98=29,4\left(g\right)\)

b, m _{dd sau pư} = 21,3 + 300 = 321,3 (g)

\(\Rightarrow C\%_{H_3PO_4}=\dfrac{29,4}{321,3}.100\%\approx9,15\%\)

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

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

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) 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)

a)\(PTHH:4P+5O_2\xrightarrow[]{t^0}2P_2O_5\)

b)\(PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\)

c)\(PTHH:2Al+3H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+3H_2\)

d)\(PTHH:Na+H_2O\xrightarrow[]{}NaO+H_2\)

e)\(PTHH:CaO+H_2O\xrightarrow[]{}Ca\left(OH\right)_2\)

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