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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.\(n_{CH_4}=\dfrac{V_{CH_4}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,3 0,6 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,6.22,4=13,44l\)
b.
\(n_P=\dfrac{m_P}{M_P}=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
0,1 0,05 ( mol )
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,05.142=7,1g\)
nAl=16,2/27= 0,6(mol)
a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3
nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)
=> V(O2,đktc)=0,45 x 22,4=10,08(l)
b) nAl2O3= nAl/2=0,6/2=0,3(mol)
=>mAl2O3=102. 0,3= 30,6(g)
c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2=2. 0,45=0,9(mol)
=>mKMnO4= 158 x 0,9= 142,2(g)
mMg = 3.6/24 = 0.15 (mol)
2Mg + O2 -to-> 2MgO
0.15__0.075____0.15
mMgO= 0.15*40 = 6 (g)
VO2 = 0.075*22.4 = 1.68 (l)
2KClO3 -to-> 2KCl + 3O2
0.05_______________0.075
mKClO3 = 0.05*122.5 = 6.125 (g)
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)
c) PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)
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)