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Độ rượu:
Độ rượu = \(\dfrac{V_{rượu}}{V_{hh}}\cdot100^o=\dfrac{2}{2+18}\cdot100^o=10^o\)
Nghĩa là cứ 100 ml rượu thì có 45 ml rượu nguyên chất
\(V_{C_2H_5OH}=\dfrac{98.45}{100}=44,1\left(ml\right)\\ V_{ddC_2H_5OH\left(25^o\right)}=\dfrac{44,1.100}{25}=176,4\left(ml\right)\)
\(v_{C_2}H_SOH\)=90:100.10=9(ml)
Độ rượu sau :
S=9/20.100=45o
\(V_{C_2H_5OH\left(nguyên.chất\right)}=\dfrac{0,5.30}{100}=0,15l\)
\(0,15lít=150ml\)
\(V_{H_2O}=500-150=350ml\)
\(m_{C_2H_5OH\left(nguyên.chất\right)}=150.0,8=120g\)
\(m_{H_2O}=350.1=350g\)
\(n_{C_2H_5OH}=\dfrac{120}{46}=2,6mol\)
\(n_{H_2O}=\dfrac{350}{18}=19,44mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
2,6 1,3 ( mol )
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
19,44 9,72 ( mol )
\(V_{H_2}=\left(1,3+9,72\right).22,4=246,848l\)
a)
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
V rượu = 57,5.12/100 = 6,9(lít) = 6900(cm3)
=> m rượu = 6900.0,8 = 5520(gam)
Theo PTHH :
n CH3COOH = n C2H5OH = 5520/46 = 120(mol)
m CH3COOH = 120.60 = 7200(gam)
b)
m dd giấm = 7200/4% = 180 000(gam)
\(V_r=57.5\cdot0.12=6.9\left(l\right)\)
\(m_{C_2H_5OH}=6.9\cdot0.8=5.52\left(g\right)\)
\(n_{C_2H_5OH}=\dfrac{5.52}{46}=0.12\left(mol\right)\)
\(n_{C_2H_5OH\left(pư\right)}=0.12\cdot92\%=0.1104\left(mol\right)\)
\(C_2H_5OH+O_2\underrightarrow{mg}CH_3COOH+H_2O\)
\(0.1104........................0.1104\)
\(m_{dd_{CH_3COOH}}=\dfrac{0.1104\cdot60}{4\%}=165.6\left(g\right)\)
a, \(V_{C_2H_5OH}=\dfrac{10.9}{100}=0,9\left(l\right)=900\left(ml\right)\)
\(\Rightarrow m_{C_2H_5OH}=900.0,8=720\left(g\right)\Rightarrow n_{C_2H_5OH}=\dfrac{720}{46}=\dfrac{360}{23}\left(mol\right)\)
PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=\dfrac{360}{23}\left(mol\right)\)
Mà: H = 92%
\(\Rightarrow n_{CH_3COOH\left(TT\right)}=\dfrac{360}{23}.92\%=14,4\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=14,4.60=864\left(g\right)\)
b, \(m_{ddgiam}=\dfrac{864}{5\%}=17280\left(l\right)\)
