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$PTHH : CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4 đ,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O \\ n_{CH_3COOH} = \dfrac{3}{60} = 0,05(mol) \\ n_{C_2H_5OH} = \dfrac{2,5}{46} = 0,054(mol) \\ Ta có : n_{C_2H_5OH} > n_{CH_3COOH} \to C_2H_5OH dư \\ n_{ CH_3COOC_2H_5 } = n_{CH_3COOH} = 0,05(mol) \\ m_{este} = 0,05.88 = 4,4(gam) \\ m_{este(tt)} = 4,4.0,9 = 3,96(gam)$
a)
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
b)
n CH3COOC2H5 = n C2H5OH = 9,2/46 = 0,2(mol)
=> m este = 0,2.88 = 17,6 gam
c)
n este = 8,8/88 = 0,1(mol)
=> n C2H5OH = n CH3COOH = 0,1/60% = 1/6 mol
=> m C2H5OH = 46 . 1/6 = 7,67(gam) ; m CH3COOH = 60 . 1/6 = 10(gam)
nC2H5OH = 8.05/46 = 0.175 (mol)
nCH3COOH = 36/60 = 0.6 (mol)
nCH3COOC2H5 = 12.32/88 = 0.14 (mol)
C2H5OH + CH3COOH <-H2SO4đ,t0-> CH3COOC2H5 + H2O
1.......................1
0.175................0.6
LTL : 0.175/1 < 0.6/1
=> CH3COOH dư
mCH3COOH (dư) = ( 0.6 - 0.175) * 60 = 25.5 (g)
nCH3COOC2H5 = nC2H5OH = 0.175 (mol)
H% = 0.14/0.175 * 100% = 80%
\(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{27,6}{46}=0,6\left(mol\right)\)
PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)
Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\), ta được C2H5OH dư nếu pư hết.
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,5\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,5.88=44\left(g\right)\)
\(\Rightarrow H=\dfrac{26,4}{44}.100\%=60\%\)
\(n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{100}{46}=\dfrac{50}{23}\left(mol\right)\)
PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H2SO4 đặc, to)
Xét tỉ lệ: \(\dfrac{1}{1}< \dfrac{\dfrac{50}{23}}{1}\), ta được C2H5OH dư.
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=1\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=1.88=88\left(g\right)\)
\(\Rightarrow H=\dfrac{55}{88}.100\%=62,5\%\)