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\(n_{CO2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Pt : \(C_6H_{12}O_6\xrightarrow[30-35^oC]{Menrượu}2C_2H_5OH+2CO_2\)
0,5 0,5
a) \(m_{C2H5OH}=0,5.46=23\left(g\right)\)
b) Pt : \(C_2H_5OH+O_2\xrightarrow[]{Mengiấm}CH_3COOH+H_2O\)
0,5 0,5
\(m_{CH3COOH\left(lt\right)}=0,5.60=30\left(g\right)\)
⇒ \(m_{CH3COOH\left(tt\right)}=30.80\%=24\left(g\right)\)
Chúc bạn học tốt
\(C_6H_{12}O_6\underrightarrow{t^o}2C_2H_5OH+2CO_2\uparrow\)(xt : men rượu )
0,5 0,5
\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{C_2H_5OH}=0,5.46=23\left(g\right)\)
\(C_2H_5OH+O_2\underrightarrow{t^o}CH_3COOH+H_2O\) (men giấm )
0,5 0,5
\(m_{CH_3COOH}=0,5.60=30\left(g\right)\)
\(m_{CH_3COOHtt}=30.80\%=24\left(g\right)\)
\(a,n_{C_6H_{12}O_6}=\dfrac{36}{180}=0,2\left(mol\right)\)
PTHH: \(C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\uparrow\)
0,2----------------->0,4----------->0,4
=> VCO2 = 0,4.22,4 = 8,96 (l)
b, mC2H5OH = 0,4.46.50% = 9,2 (g)
\(c,V_{C_2H_5OH}=\dfrac{9,2}{0,8}=11,5\left(ml\right)\\ \rightarrow V_{ddC_2H_5OH}=\dfrac{11,5.100}{60}=\dfrac{115}{6}\left(ml\right)\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: \(C_6H_{12}O_6\dfrac{_{memruou}}{30^0-35^0C}>2C_2H_5OH+2CO_2\uparrow\)
0,25 0,25 0,25 (mol)
a) \(m_{C_2H_5OH}=0,25.46=11,5\left(g\right)\)
b) \(m_{C_6H_{12}O_6}=0,25.180.90\%=40,5\left(g\right)\)
a)
$C_6H_{12}O_6 \xrightarrow{t^o,xt} 2CO_2 + 2C_2H_5OH$
720 ml = 720 cm3
m dd glucozo = D.V = 720.1 = 720(gam)
m glucozo = 720.5% = 36(gam)
n glucozo = 36/180 = 0,2(mol)
Theo PTHH :
n C2H5OH = 2n glucozo = 0,4(mol)
m C2H5OH = 0,4.46 = 18,4(gam)
b)
V rượu = m/D = 18,4/0,8 = 23(ml)
Vậy :
Đr = 23/240 .100 = 9,583o
a, n\(C_2H_6O\)= \(\frac{6,9}{46}\)=0,15mol
pt : C2H6O + 3O2 → 2CO2 + 3 H2O
(mol) 0,15mol → 0,45mol→0,3mol
m\(CO_2\) = 0,3 . 44 = 13,2 G
b, Vkk = \(\frac{20}{100}\) : (0,45 . 22,4 ) = 50,4l
a, \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
Theo PT: \(n_{CO_2}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow V_{CO_2}=\dfrac{2}{15}.22,4=\dfrac{224}{75}\left(l\right)\)
b, \(n_{C_2H_6O\left(LT\right)}=\dfrac{1}{3}n_{O_2}=\dfrac{1}{15}\left(mol\right)\)
Mà: H = 90%
\(\Rightarrow n_{C_2H_6O\left(TT\right)}=\dfrac{\dfrac{1}{15}}{90\%}=\dfrac{2}{27}\left(mol\right)\)
\(\Rightarrow m_{C_2H_6O}=\dfrac{2}{27}.46=\dfrac{92}{27}\left(g\right)\)
\(n_{C_6H_{12}O_6}=\dfrac{18}{180}=0.1\left(mol\right)\)
\(C_6H_{12}O_6\underrightarrow{mr}2C_2H_5OH+2CO_2\)
\(0.1.....................0.2...............0.2\)
\(n_{CaCO_3}=n_{CO_2}=\dfrac{12.5}{100}=0.125\left(mol\right)\)
\(H\%=\dfrac{0.125}{0.2}\cdot100\%=62.5\%\)
\(n_{C_2H_5OH}=0.125\left(mol\right)\)
\(m_{C_2H_5OH}=0.125\cdot46=5.75\left(g\right)\)
