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a, PTHH: 4P + 5O2 \(\underrightarrow{t^o}\) 2P2O5
b, Theo ĐLBTKL, ta có:
mP + mO\(_2\) = m\(P_2O_5\)
=> mP = 28,4 - 16 = 12,4 (g )
\(PTHH:4P+5O_2\rightarrow2P_2O_5\)
a) \(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)
Theo PTHH:
\(n_{P2O5}=0,5.n_P=0,5.0,2=0,1\left(mol\right)\)
\(m_{P2O5}=0,1.142=14,2\left(g\right)\)
b) \(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PTHH:
\(n_{P2O5}=\frac{2}{5}.n_{O2}=0,2\left(mol\right)\)
\(m_{P2O5}=142.0,2=28,4\left(g\right)\)
c) \(n_P=12,431=0,4\left(mol\right)\)
\(n_{O2}=\frac{17}{32}=0,53125\left(mol\right)\)
Lập tỉ lệ: \(\frac{0,4}{4}< \frac{0,53125}{5}\)
Nên O2 dư, P hết
Theo PTHH:
\(n_{P2O5}=0,5n_P=0,5.0,4=0,2\left(mol\right)\)
\(m_{P2O5}=142.0,2=28,4\left(g\right)\)
e 1 )
\(n_{O2\left(dư\right)}=0,53125-0,5=0,03125\left(mol\right)\)
\(V_{O2\left(Dư\right)}=22,4.0,03125=0,7\left(mol\right)\)
d) \(n_P=\frac{15,5}{31}=0,5\left(mol\right)\)
\(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Tỉ lệ: \(\frac{0,5}{4}>\frac{0,5}{5}\)
Nên P dư, O2 hết
Theo PTHH: \(n_{P2O5}=\frac{2}{5}n_{O2}=0,2\left(mol\right)\)
\(\Rightarrow m_{P2O5}=142.0,2=28,4\left(g\right)\)
e2)
\(n_{P\left(dư\right)}=0,5.\left(\frac{4}{5}.0,5\right)=0,1\left(mol\right)\)
\(m_{P\left(dư\right)}=31.0,1=3,1\left(g\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
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: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
1. \(4P+5O_2\underrightarrow{^{t^o}}2P_2O_5\)
2. Ta có: \(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,02\left(mol\right)\Rightarrow m_{P_2O_5}=0,02.142=2,84\left(g\right)\)
3. \(n_{O_2}=\dfrac{5}{4}n_P=0,05\left(mol\right)\Rightarrow V_{O_2}=0,05.22,4=1,12\left(l\right)\)
a) 4P + 5O2 --to--> 2P2O5
b) \(n_{P_2O_5}=\dfrac{34,08}{142}=0,24\left(mol\right)\)
4P + 5O2 --to--> 2P2O5
0,48<-0,6<------0,24
=> mO2 = 0,6.32 = 19,2 (g)
c)
C1: mP = 0,48.31 = 14,88(g)
C2:
Theo ĐLBTKL: mP + mO2 = mP2O5
=> mP = 34,08-19,2 = 14,88(g)
d)
VO2 = 0,6.22,4 = 13,44 (l)
=> Vkk = 13,44 :20% = 67,2 (l)
4P + 5O2 ----> 2P2O5
0,24 -> 0,3 ---> 0,12 (mol)
nP = \(\dfrac{7,44}{31}\)= 0,24 (mol)
VH2 = 0,3 . 22,4 = 6,72 (l)
2KClO3 ---> 2KCl + 3O2
0,2 <------------- 0,3 (mol)
mKClO3 = 0,2 . (39 + 35,5 + 16.3)
= 24,5 (g)
Vui lòng kiểm tra lại kết quả dùm, thank you.
nP = 7,44 : 31 = 0,24 ( mol)
pthh : 4P + 5O2 -t--> 2P2O5
0,24->0,3 (mol)
=> VO2 =0,3 . 22,4 = 6,72 (l)
pthh : 2KClO3 -t--> 2KCl + 3O2
0,2<-------------------0,3 (mol)
=> mKClO3 = 0,2 .122,5 = 24,5 (g)
\(n_{P2O5}=\frac{35,5}{142}=0,25\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^O}2P_2O_5\)
0,5___0,625___0,25(mol)
\(x=0,5.31=15,5\left(g\right)\)
\(y=0,625.22,4=14\left(l\right)\)
\(\Rightarrow ChọnđápánC\)