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\(D=0,8g\)/cm3
Trong 200ml rượu etylic \(11,5^o\) có:
\(V_{C_2H_5OH}=V_{dd}\cdot\dfrac{11,5}{100}=200\cdot\dfrac{11,5}{100}=23ml\)
\(\Rightarrow m_{C_2H_5OH}=D\cdot V=23\cdot0,8=18,4g\)
\(C_2H_5OH+O_2\rightarrow CH_3COOH+H_2O\)
46 60 (gam)
18,4 m (gam)
\(\Rightarrow m=24g\)
\(m_{ddCH_3COOH}=\dfrac{24}{15\%}\cdot100\%=160g\)
KLR của rượu etylic là 0,8 (g/cm3)
a/ \(n_{CH_3COOH}=0,15.2=0,3\left(mol\right)\)
PTHH: C2H5OH + CH3COOH ----H2SO4 đ,to→ CH3COOC2H5 + H2O
Mol: 0,3 0,3
\(m_{C_2H_6O}=0,3.46=13,8\left(g\right)\Rightarrow V_{C_2H_6O}=\dfrac{13,8}{0,8}=17,25\left(ml\right)\)
\(\RightarrowĐộ.rượu=\dfrac{17,25}{500}.100\%=3,45^o\)
b,
PTHH: 2C2H5OH + 2K → 2C2H5OK + H2
Mol: 0,3 0,3
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)
a, \(C_{M_{CH_3COOH}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)
nNaOH=0,3(mol)
CH3COOH + NaOH -> CH3COONa + H2O
x__________x__________________x(mol)
CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH
y_____________y(mol)
Hệ pt:
\(\left\{{}\begin{matrix}60x+88y=20,8\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
mH2O=18x=18.0,2=3,6(g) => V(H2O)=3,6(ml)
mC2H5OH=46y=46.0,1=4,6(g) => V(H2O)= 46/0,8=57,5(ml)
=> \(D_r=\dfrac{57,5}{57,5+3,6}.100\approx94,1^o\)
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{2,84}{142}=0,02\left(mol\right)\)
PTHH :
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
0,04 0,02 0,02
\(a,C_M=\dfrac{n}{V}=\dfrac{0,04}{0,1}=0,4M\)
\(b,V_{H_2}=0,02.22,4=0,448\left(l\right)\)
\(c,PTHH:\)
\(CH_3COOH+C_2H_5OH\underrightarrow{t^o,H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
0,04 0,04
\(m_{este}=0,04.90\%.88=3,168\left(g\right)\)
\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)
\(n_{CH_3COOH}=\dfrac{300\cdot5\%}{60}=0.25\left(mol\right)\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(0.25........................................................0.125\)
\(V_{H_2}=0.125\cdot22.4=2.8\left(l\right)\)
\(n_{C_2H_5OH}=0.1\cdot2=0.2\left(mol\right)\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(ĐK:H_2SO_{4\left(đ\right)},t^0\right)\)
\(0.2......................0.2.....................0.2\)
\(\Rightarrow CH_3COOHdư\)
\(m_{CH_3COOC_2H_5}=0.2\cdot88=17.6\left(g\right)\)
Bài 1:
nCH3COOH = 0,08.1,5 = 0,12 (mol)
PTHH: CH3COOH + C2H5OH --H+,to--> CH3COOC2H5 + H2O
0,12----------------------------->0,12
=> mCH3COOC2H5 = 0,12.88 = 10,56 (g)
Bài 2:
nCH3COOH = 2.0,1 = 0,2 (mol)
PTHH: 2CH3COOH + Mg --> (CH3COO)2Mg + H2
0,2------->0,1----------------------->0,1
=> mMg = 0,1.24 = 2,4 (g)
PTHH: C2H4 + H2 --to,Ni--> C2H6
0,1<--0,1
=> VC2H4(đktc) = 0,1.22,4 = 2,24 (l)