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a, C\(_2\)H\(_5\)OH + O\(_2\) →(t\(^0\)) CH\(_3\)COOH + H\(_2\)O
b, V\(_r\) = \(\frac{500\cdot10}{100}\) = 50(ml)
V\(_{nc}\) = V\(_{hh}\) - V\(_r\) = 500 - 50 = 450 (ml)
⇒ m\(_{nc}\) = D\(_{nc}\) * V\(_{nc}\) = 450 * 1 = 450 (gam)
m\(_r\) = D\(_r\) * V\(_r\) = 0,78 * 50 = 39 (gam)
C\(_2\)H\(_5\)OH + O\(_2\) →(t\(^0\)) CH\(_3\)COOH + H\(_2\)O
(gam) 46 32 81
đề (gam pư) : (39*80%) → y → x
⇒ m\(_{CH_3COOH_{\left(tte\right)}}\) = x = \(\frac{39\cdot80\left(\%\right)\cdot81}{46}\) \(\sim\) 54, 94(gam)
m\(_{rượu}còn\) = 39 - (39* 80%) = 7,8 (gam)
m\(_{O_2\left(PƯ\right)}\) = y = \(\frac{39\cdot\left(80\%\right)\cdot32}{46}\) \(\sim\) 21,7 (gam)
m\(_{ddsau}\) = m\(_{dbđ}\) + m\(_{O_2\left(PƯ\right)}\) = m \(_r\) + m\(_{nc}\) + m\(_{O_2\left(PƯ\right)}\) = 39 + 450 + 21,7 = 510,7 (gam)
dung dịch sau pư có axit axetic và rượu etylic dư
C\(_{\%_rcòn}\) = \(\frac{7,8\cdot100\%}{510,7}\) \(\sim\) 1,53 (%)
C\(_{\%_{CH_3COOH}}\) = \(\frac{54,94\cdot100\%}{510,7}\) \(\sim\) 10,76(%)
a) n glucozo = 54/180 = 0,3(mol)
n glucozo pư = 0,3.80% = 0,24(mol)
$C_6H_{12}O_6 \xrightarrow{t^o} 2CO_2 +2 C_2H_5OH$
n C2H5OH = 2n glucozo = 0,48(mol)
m C2H5OH = 0,48.46 = 22,08(gam)
b)
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
n CH3COOH = n C2H5OH = 0,48(mol)
C% CH3COOH = 0,48.60/500 .100% = 5,76%
Bài 1:
PTHH: \(C_2H_5OH+O_2\xrightarrow[]{mengiấm}CH_3COOH+H_2O\)
Ta có: \(n_{C_2H_5OH}=\dfrac{115\cdot0,8}{46}=2\left(mol\right)=n_{CH_3COOH\left(lýthuyết\right)}\)
\(\Rightarrow m_{CH_3COOH\left(thực\right)}=2\cdot60\cdot90\%=108\left(g\right)\)
Bài 2:
PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4\left(đ\right)]{t^o}CH_3COOC_2H_5+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{92}{46}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Rượu còn dư, Axit p/ứ hết
\(\Rightarrow n_{CH_3COOC_2H_5\left(lýthuyết\right)}=1\left(mol\right)\) \(\Rightarrow m_{CH_3COOC_2H_5\left(thực\right)}=1\cdot88\cdot80\%=70,4\left(g\right)\)
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)
Đổi 10kg = 10000g
Ta có: \(n_{CH_3COOH\left(LT\right)}=\dfrac{10000.5\%}{92\%}=\dfrac{12500}{23}\left(mol\right)\)
PTHH:
\(C_2H_5OH+O_2\xrightarrow[]{\text{men giấm}}CH_3COOH+H_2O\)
\(\dfrac{12500}{23}\)<---------------------\(\dfrac{12500}{23}\)
\(\Rightarrow m_{C_2H_5OH}=\dfrac{12500}{23}.46=25000\left(g\right)=25\left(kg\right)\)
