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Câu 7.
a. \(n_P=\dfrac{15.5}{31}=0,5\left(mol\right)\)
PTHH : 4P + 5O2 ----to---> 2P2O5
0,5 0,625 0,25
\(m_{P_2O_5}=0,25.142=35,5\left(g\right)\)
b. \(V_{O_2}=0,625.22,4=14\left(l\right)\\ \Rightarrow V_{kk}=14.5=70\left(l\right)\)
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)
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: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
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)
a) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
0,3--->0,2----->0,1
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)
c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)
\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)
PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)
0,44<------------------------------------0,22
\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)
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