Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
4P+5O2-to>2P2O5
0,04----0,05----0,02
n P=0,04 mol
=>m P2O5=0,02.142=2,84g
=>VO2=0,05.22,4=1,12l
c)
2KMnO4-to>K2MnO4+MnO2+O2
0,1----------------------------------------0,05
H=10%
m KMnO4=0,1.158.110%=17,28g
\(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{T^O}2P_2O_5\)
0,04 0,05 0,02
=> \(\left\{{}\begin{matrix}m_{P_2O_5}=0,02.142=2,84\left(g\right)\\V_{O_2}=0,05.22,4=1,12\left(l\right)\end{matrix}\right.\)
\(pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
=> \(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
\(m_{KMnO_4\left(d\text{ùng}\right)}=15,8.110\%=17,38\left(g\right)\)
nP = 12,4/31 = 0,4 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
Mol: 0,4 ---> 0,5 ---> 0,2
mP2O5 = 0,2 . 142 = 28,4 (g)
VO2 = 0,5 . 22,4 = 11,2 (l)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
nKMnO4 = 0,5 . 2 = 1 (mol)
mKMnO4 = 1 . 158 = 158 (g)
Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
____0,4____0,5____0,2 (mol)
a, \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
b, \(V_{O_2}=0,5.22,4=11,2\left(l\right)\Rightarrow V_{kk}=5V_{O_2}=56\left(l\right)\)
nP= 12.4/31=0.4 mol
4P + 5O2 -to-> 2P2O5
0.4___0.5
VO2= 0.5*22.4=11.2l
VKK=5VO2= 11.2*5=56l
2KClO3 -to-> 2KCl + 3O2
1/3________________0.5
mKClO3= 1/3*122.5=245/6g
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,6 0,3 0,6 ( mol )
\(m_{H_2O}=0,6.18=10,8g\)
\(V_{kk}=V_{O_2}.5=\left(0,3.22,4\right).5=33,6l\)
a, \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=11,2\left(l\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
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: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
a) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,2-->0,25
=> VO2 = 0,25.22,4 = 5,6 (l)
=> Vkk = 5,6.5 = 28 (l)
b)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,5<-----------------------------0,25
=> \(m_{KMnO_4}=0,5.158=79\left(g\right)\)
a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,4-->0,5------->0,2
=> \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
b) \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
c) Vkk = 11,2.5 = 56 (l)
lấy đâu ra 31 thế ạ