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\(V_{C_2H_5OH}=\dfrac{45.23}{100}=10,35\left(ml\right)\)
=> \(m_{C_2H_5OH}=10,35.0,8=8,28\left(g\right)\)
=> B
nC2H5OH=0,5(mol)
V(C2H5OH)=23/0,8=28,75(ml)
=> A=Dr= (28,75/250).100=11,5o
PTHH: C2H5OH + O2 -men giấm---> CH3COOH + H2O
nCH3COOH=C2H5OH=0,5(mol)
=>mCH3COOH=0,5. 60=30(g)
=> m(giấm ăn)= 30/5%=600(g)
=>a=600(g)
a)
\(V_{C_2H_5OH}=\dfrac{96.20}{100}=19,2\left(ml\right)\)
=> \(m_{C_2H_5OH}=19,2.0,8=15,36\left(g\right)\)
b) \(n_{C_2H_5OH}=\dfrac{15,36}{46}=\dfrac{192}{575}\left(mol\right)\)
\(V_{H_2O}=20-19,2=0,8\left(ml\right)\)
=> \(m_{H_2O}=0,8.1=0,8\left(g\right)\)
=> \(n_{H_2O}=\dfrac{0,8}{18}=\dfrac{2}{45}\left(mol\right)\)
PTHH: 2C2H5OH + 2Na --> 2C2H5ONa + H2
\(\dfrac{192}{575}\)------------------------->\(\dfrac{96}{575}\)
2H2O + 2Na --> 2NaOH + H2
\(\dfrac{2}{45}\)----------------------->\(\dfrac{1}{45}\)
=> \(V_{H_2}=22,4.\left(\dfrac{96}{575}+\dfrac{1}{45}\right)=4,238\left(l\right)\)
C2H5OH + Na -- > C2H5OHNa + 1/2 H2
Na+H2O --- > NaOH + 1/2H2
Vr = 20x96/100 = 19,2ml = 0.0192 (l)
mC2H5OH = D.V = 19,2 x 0.8 = 15.36 (g)
nC2H5OH = m/M = 15.36 / 46 = 0.43 (mol)
=> nH2 = 0.215 (mol)
VH2O = 1 ml => mH2O = 1 (g)
=> nH2O = m/M = 1/18 = 0.056 (mol)
=> nH2 = 0.028 (mol)
nH2 = 0.215 + 0.028 = 0.243 (mol)
=> VH2 = 22.4 x 0,243 = 5,4432 (l)
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: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5ONa}=0,4\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)
b1
=>Đr=50\150.100=33,30