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n CH4 = 1.85% = 0,85(mol)
n C2H6 = 1.10% = 0,1(mol)
$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$C_2H_6 + \dfrac{7}{2} O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
Theo PTHH :
n O2 = 2n CH4 + 7/2 n C2H6 = 2,05(mol)
n không khí = n O2 : 20% = 2,05 : 20% = 10,25(mol)
\(m_{CH_4}=0,3.16=4,8(g)\)
Bảo toàn KL: \(m_{CH_4}+m_{O_2}=m_{CO_2}+m_{H_2O}\)
\(\Rightarrow m_{O_2}=10,8+13,2-4,8=19,2(g)\\ \Rightarrow V_{O_2}=\dfrac{19,2}{32}.22,4=13,44(l)\\ \Rightarrow V_{kk}=13,44.5=67,2(l)\)
\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\\ PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
tỉ lệ 4 : 5 : 2
n(mol) 0,2--->0,25-------->0,1
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\\ V_{kk}=5,6:\dfrac{1}{5}=28\left(l\right)\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=28\left(l\right)\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\\ n_{O_2}=2.n_{CH_4}=2.0,2=0,4\left(mol\right)\\ a,V_{kk}=5.V_{O_2\left(đktc\right)}=5.\left(0,4.22,4\right)=44,8\left(l\right)\\ b,m_{CO_2}=0,2.44=8,8\left(g\right)\)
nCH4 = 6,72/22,4 = 0,3 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,3 ---> 0,6
Vkk = 0,6 . 22,4 : 21% = 64 (l)
-PTHH: \(4P+5O_2\rightarrow^{t^0}2P_2O_5\).
-\(n_P=\dfrac{m}{M}=\dfrac{3,1}{31}=0,1\left(mol\right)\)
-Theo PTHH trên, ta có:
-\(n_{O_2}=\dfrac{0,1}{4}.5=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=n.22,4=0,125.22,4=2,8\left(l\right)\)
\(\Rightarrow V_{KK}=V_{O_2}.5=2,8.5=14\left(l\right)\)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ n_{O_2}=\dfrac{5}{4}.0,1=0,125\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\\ V_{kk\left(đktc\right)}=2,8.5=14\left(l\right)\)
PTHH: 2Cu + O2 ---to→ 2CuO
Mol: x 0,5x
PTHH: 2Mg + O2 ---to→ 2MgO
Mol: 0,5x 0,25x (do số mol của Cu gấp đôi Mg)
Ta có: \(64x+24.0,25x=15,2\Leftrightarrow x=\dfrac{38}{175}\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,5.\dfrac{38}{175}+0,25.\dfrac{38}{175}=\dfrac{57}{350}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{57}{350}.22,4=3,648\left(l\right)\)
\(\Rightarrow V_{kk}=3,648.5=18,24\left(l\right)\)
a) nAl=0,2(mol)
PTHH: 4Al +3 O2 -to-> 2 Al2O3
0,2_______0,15___0,1(mol)
V(O2,đktc)=0,15.22,4=3,36(l)
V(kk)=5.V(O2,đktc)=3,36.5=16,8(l)
b) mAl2O3=0,1.102=10,2(g)
a) \(CH_4+2O_2\rightarrow CO_2+2H_2O\)
0,1 0,2
\(V_{O_2}=0,2\cdot22,4=4,48l\)
\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot4,48=22,4l\)
b) \(C_4H_{10}+\dfrac{13}{2}O_2\rightarrow4CO_2+5H_2O\)
0,1 0,65
\(V_{O_2}=0,65\cdot22,4=14,56l\)
\(V_{kk}=5V_{O_2}=5\cdot14,56=72,8l\)
Dùng \(0,1molC_4H_{10}\) cần nhiều lượng không khí hơn.