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\(n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\a, PTHH:C_2H_5OH+3O_2\rightarrow\left(t^o\right)2CO_2+3H_2O\\ b,n_{O_2}=3.0,5=1,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=22,4.1,5=33,6\left(l\right)\\ c,V_{C_2H_5OH}=46\%.100=46\left(ml\right)\\ C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\\ n_{C_2H_5OH}=\dfrac{0,8.46}{46}=0,8\left(mol\right)\\ n_{H_2}=\dfrac{0,8}{2}=0,4\left(mol\right)\Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\)
a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
b, \(n_{C_2H_6O}=\dfrac{23}{46}=0,5\left(mol\right)\)
Theo PT: \(n_{O_2}=3n_{C_2H_6O}=1,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)
c, \(V_{C_2H_6O}=\dfrac{100.46}{100}=46\left(ml\right)\)
\(\Rightarrow m_{C_2H_6O}=46.0,8=36,8\left(g\right)\)
\(\Rightarrow n_{C_2H_6O}=\dfrac{36,8}{46}=0,8\left(mol\right)\)
PT: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5ONa}=0,4\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)
a) C2H5OH + 3O2 --to--> 2CO2 + 3H2O
b) \(n_{C_2H_5OH}=\dfrac{2,3}{46}=0,05\left(mol\right)\)
PTHH: C2H5OH + 3O2 --to--> 2CO2 + 3H2O
0,05-->0,15-------->0,1
=> VCO2 = 0,1.22,4 = 2,24 (l)
c) VO2 = 0,15.22,4 = 3,36 (l)
=> Vkk = 3,36 : 20% = 16,8 (l)
\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)
0,1 0,3 0,2
b) \(m_{C2H5OH}=0,1.46=4,6\left(g\right)\)
c) \(V_{O2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
Chúc bạn học tốt
TK
Từ C2H4O2 ta có: M = 60 g/mol; mC = 2 x 12 = 24 g; mH = 4 x 1 = 4 g;
MO = 2 x 16 = 32 g.
%C = (24 : 60) x 100% = 40%; %H = (4 : 60) x 100% = 6,67%;
%O = 100% - 40% - 6,67% = 53,33%.
a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
b, \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{C_2H_6O}=0,1.46=4,6\left(g\right)\)
\(\Rightarrow V_{C_2H_6O}=\dfrac{4,6}{0,8}=5,75\left(ml\right)\)
Độ rượu = \(\dfrac{5,75}{50}.100=11,5^o\)
\(n_{C_2H_6O}=\dfrac{23}{46}=0,5\left(mol\right)\)
PTHH: C2H6O + 3O2 --to--> 2CO2 + 3H2O
0,5------------------->1
=> VCO2 = 1.22,4 = 22,4 (l)
a) PTHH : \(C_2H_6O+3O_2\left(t^o\right)->2CO_2+3H_2O\) (1)
b) \(n_{C_2H_6O}=\dfrac{m}{M}=\dfrac{23}{12.2+1.6+16}=0,5\left(mol\right)\)
Từ (1) -> \(n_{CO_2}=2n_{C_2H_6O}=1\left(mol\right)\)
=> \(V_{CO_2\left(đktc\right)}=n.22,4=1.22,4=22,4\left(l\right)\)
a) $C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
b) $n_{C_2H_5OH} = \dfrac{4,6}{46} = 0,1(mol)$
$n_{O_2} = 3n_{C_2H_5OH} = 0,3(mol)$
$V_{O_2} = 0,3.22,4 = 6,72(lít)$
c)
Theo PTHH :
$n_{CO_2} = 2n_{C_2H_5OH} = 0,2(mol) \Rightarrow V_{CO_2} = 0,2.22,4 = 4,48(lít)$
$n_{H_2O} = 3n_{C_2H_5OH} = 0,3(mol) \Rightarrow m_{H_2O} = 0,3.18 = 5,4(gam)$
