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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
\(m_{Al_2O_3}=\dfrac{2}{15}.102=13,6\left(g\right)\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,6 0,3 0,6 ( mol )
\(m_{H_2O}=0,6.18=10,8g\)
\(V_{kk}=V_{O_2}.5=\left(0,3.22,4\right).5=33,6l\)
\(n_{H_2}\)=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH 2H2 +O2----to--->2H2O
0,2....0,1.................0,2
=>\(m_{H_2O}=0,2.18=3,6\left(g\right)\)
=>\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
=>Vkk=2,24.5=11,2(l)
\(n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{H_2O} = n_{H_2} =0,2(mol) \Rightarrow m_{H_2O} = 0,2.18 = 3,6(gam)\\ n_{O_2} = \dfrac{1}{2}n_{H_2} = 0,1(mol)\\ \Rightarrow V_{O_2} = 0,1.22,4 = 2,24(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 2,24.5 = 11,2(lít) \)
a)\(2Mg + O_2 \xrightarrow{t^o} 2MgO\)
b)
\(n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\)
Theo PTHH :
\(n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,05(mol)\\ \Rightarrow V_{O_2} = 0,05.22,4 = 1,12(lít)\)
c)
\(n_{MgO} = n_{Mg} = 0,1(mol)\\ \Rightarrow m_{MgO} = 0,1.40 = 4(gam)\)
d)
\(V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)
nAl = 2,7/27 = 0,1 (mol)
PTHH: 4Al + 3O2 -> (t°) 2Al2O3
Mol: 0,1 ---> 0,075 ---> 0,05
mAl2O3 = 0,05 . 102 = 5,1 (g)
VO2 = 0,075 . 22,4 = 1,68 (l)
Vkk = 1,68 . 5 = 8,4 (l)
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
0,1 0,075 0,05 ( mol )
\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,05.102=5,1g\)
\(V_{kk}=V_{O_2}.5=\left(0,075.22,4\right).5=8,4l\)
a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,15<---0,3<----0,15
b) `m_{O_2} = 0,3.32 = 9,6 (g)`
c) `V_{CH_4} = 0,15.22,4 = 3,36 (l)`
nSO2 = 12,8 : 64=0,2 (mol)
pthh : S+ O2 -t->SO2
0,2<--0,2<------0,2(mol)
=> mS= 0,2.32=6,4 (g)
=> VO2= 0,2.22,4=4,48 (l)
ta có
VO2 = 1/5 Vkk <=> Vkk = VO2 : 1/5 = 4,48:1/5 = 22.4 (l)
S + O2 to→to→ SO2
nS=12,832=0,4(mol)
a) Theo PT: nSO2=nS=0,4(mol)
⇒VSO2=0,4×22,4=8,96(l)
b) Theo PT: nO2=nS=0,4(mol)
⇒VO2=0,4×22,4=8,96(l)
⇒VKK=5VO2=5×8,96=44,8(l)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
b, \(n_{CH_4}=\dfrac{28}{22,4}=1,25\left(mol\right)\)
\(n_{CO_2}=n_{CH_4}=1,25\left(mol\right)\Rightarrow m_{CO_2}=1,25.44=55\left(g\right)\)
c, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\)
Câu 1 :
\(a,2KMnO_4\underrightarrow{^{to}}K_2MnO_4+MnO_2+O_2\uparrow\)
Ta có : \(n_{KMnO4}=\frac{31,6}{158}=0,2\left(mol\right)\)
\(\rightarrow n_{O2}=\frac{1}{2}n_{KMnO4}=0,1\left(mol\right)\)
\(\rightarrow V_{O2}=0,1.22,4=2,24\left(l\right)\)
\(b,2Cu+O_2\underrightarrow{^{to}}2CuO\)
Ta có : \(n_{O2}=0,1\left(mol\right)\rightarrow n_{CuO}=2n_{O2}=0,2\left(mol\right)\)
\(\rightarrow m_{CuO}=0,2.50=16\left(g\right)\)
Câu 2:
\(a,CH_4+2O_2\underrightarrow{^{to}}CO_2+2H_2O\)
b, Ta có :
\(n_{CH4}=\frac{17,92}{22,4}=0,8\left(mol\right)\)
\(n_{H2O}=2n_{CH4}=1,6\left(mol\right)\rightarrow m_{H2O}=1,6.18=28,8\left(g\right)\)
c, \(n_{O2}=2n_{CH4}=1,6\left(mol\right)\)
\(\rightarrow n_{kk}=5n_{O2}=8\left(mol\right)\)
\(\rightarrow V_{kk}=8.22,4=179,2\left(l\right)\)
Câu 1 :
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
Câu 2 :
nCH4 = 17.92/22.4 = 0.8 mol
CH4 + 2O2 -to-> CO2 + 2H2O
0.8_____1.6______0.8____1.6
mH2O = 1.6*18 = 28.8 g
VCO2 = 0.8*22.4 = 17.92 l
Vkk = 5VO2 = 5*1.6*22.4 = 168 (l)