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K 2 MnO 4 + MnO 2 + O 2 . 
nKMnO4 = 63,2 : 158 = 0,4( mol)
pthh : 2MKMnO4 -t--> K2MnO4 + MnO2 + O2
0,4 --------------------------------------->0,2 (mol)
-> VO2 = 0,2 .22,4 = 4,48 (L)
PTHH : 4Al + 3O2 -t--> 2Al2O3
0 ,2 -----> 2/15 (mol)
=> mAl2O3 = 2/15 . 102 = 13,6 (g)
a, \(n_{KMnO_4}=\dfrac{63,2}{158}=0,4\left(mol\right)\)
PTHH: 2KMnO4 ---to→ K2MnO4 + MnO2 + O2
Mol: 0,4 0,2
\(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
b,
PTHH: 4Al + 3O2 ---to→ 2Al2O3
Mol: 0,2 0,133
\(m_{Al_2O_3}=\dfrac{2}{15}.102=13,6\left(g\right)\)
nO2 = 2,24/22,4 = 0,1 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
nKMnO4 = 0,1 . 2 = 0,2 (mol)
mKMnO4 = 0,2 . 158 = 31,6 (g(
Để tạo hh nổ mạnh nhất thì VH2 : VO2 = 2 : 1
=> VH2 = VO2 . 2 = 2,24 . 2 = 4,48 (l)
PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
a+b) Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
\(\Rightarrow n_{MnO_2}=n_{O_2}=0,1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{MnO_2}=0,1\cdot87=8,7\left(g\right)\\V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
c) PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Theo PTHH: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,05\cdot232=11,6\left(g\right)\)
a, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)
Theo PT: \(n_{K_2MnO_4}=\dfrac{1}{2}n_{KMnO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{K_2MnO_4}=0,1.197=19,7\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.24,79=2,479\left(l\right)\)
c, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,05\left(mol\right)\\n_{H_2O}=n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{CO_2}=0,05.24,79=1,2395\left(l\right)\)
\(m_{H_2O}=0,1.18=1,8\left(g\right)\)
a)nO2=\(\dfrac{3.36}{22.4}\)=0,15(mol)
2KMnO4(to)→K2MnO4+MnO2+O2
Theo PT: nKMnO4=2nO2=0,3(mol)
→m=mKMnO4=0,3.158=47,4(g)
b)nH2=\(\dfrac{8.96}{22.4}\)=0,4(mol)
2H2+O2(to)→2H2O
Vì \(\dfrac{nH_2}{2}\)<nO2→O2nH2 dư
Theo PT: nH2O=nH2=0,4(mol)
→mH2O=0,4.18=7,2(g)
a.b.\(n_{KMnO_4}=\dfrac{m}{M}=\dfrac{31,6}{158}=0,2mol\)
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,2 0,1 ( mol )
\(V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24l\)
c.\(3Fe+2O_2\rightarrow Fe_3O_4\)
0,1 0,05 ( mol )
\(m_{Fe_3O_4}=n_{Fe_3O_4}.M_{Fe_3O_4}=0,05.232=11,6g\)
a) \(n_{KMnO_4}=\frac{31,6}{158}=0,2\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,2--------------0,1--------------------------->(mol)
\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
b) nO2=0,1(mol)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
0,2------0,1----0,2(mol)
\(m_{Cuo}=0,2.64=12,8\left(g\right)\)
nKMnO4 = 31.6/158 = 0.2 mol
2KMnO4 -to-> K2MnO4 + MnO2 + O2
0.2____________________________0.1
VO2 = 0.1*22.4 = 2.24 (l)
Cu + 1/2O2 -to-> CuO
_______0.1______0.2
mCuO = 0.2*80 = 16 g