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mCH3COOH = 12g
nCH3COOH = 0.2 mol
CH3COOH + NaHCO3 --> CH3COONa + CO2 + H2O
0.2____________0.2___________0.2_______0.2
mNaHCO3 = 16.8 g
mdd NaHCO3 = 200 g
mCH3COONa = 16.4 g
mCO2 = 8.8 g
mdd sau phản ứng = 100 + 200 - 8.8 = 291.2 g
C%CH3COONa = 16.4/291.2*100% = 5.63%
PTHH.
CH3COOH + NaHCO3 -> CH3COONa + CO2 + H2O
0,2....................0,2..................0,2...............0,2.........0,2 (mol)
Theo bài:
mCH3COOH = \(\frac{100.12}{100}\)= 12 g
=> nCH3COOH = 12/60 = 0,2
Theo pthh và bài có:
nNaHCO3 = nCH3COOH = 0,2 mol
=> mNaHCO3 = 0,2 . 84 = 16,8 g
mddNaHCO3=\(\frac{16,8.100}{8,4}=200\left(g\right)\)
+nCH3COONa = nCH3COOH = 0,2 mol
=> mCH3COONa = 0,2.82 = 16,4 (g)
m dd sau pư = mddCH3COOH + mddNaHCO3− mCO2
= 100 + 200 - 0,2.44= 291,2 (g)
=> C%dd CH3COONa = \(\frac{16,4.100}{291,2}=5,63\left(\%\right)\)
vậy....
a) PTHH:\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
b) Ta có: \(n_{CaCO_3}=\frac{60}{100}=0,6\left(mol\right)\) \(\Rightarrow n_{CH_3COOH}=1,2mol\)
\(\Rightarrow m_{CH_3COOH}=1,2\cdot60=72\left(g\right)\) \(\Rightarrow m_{ddCH_3COOH}=\frac{72}{12\%}=600\left(g\right)\)
c) Theo PTHH: \(n_{CaCO_3}=n_{\left(CH_3COO\right)_2Ca}=n_{CO_2}=0,6mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Ca}=0,6\cdot158=94,8\left(g\right)\\m_{CO_2}=0,6\cdot44=26,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{CaCO_3}+m_{ddCH_3COOH}-m_{CO_2}=633,6\left(g\right)\)
\(\Rightarrow C\%_{dd\left(CH_3COO\right)_2Ca}=\frac{94,8}{633,6}\cdot100\approx14,96\%\)
d) Ta có: \(n_{CO_2}=n_{BaCO_3}=0,6mol\)
\(\Rightarrow m_{BaCO_3}=0,6\cdot197=118,2\left(g\right)\)
2
.Zn+2CH3COOH-->(CH3COO)2Zn+H2
0,1--------------------------------------------0,1 mol
nZn=6,5\65=0,1 mol
=>VH2=0,1.22,4=2,24 g
1.
CH3COOH+C2H5OH--->CH3COOC2H5+h2o
0,5-------------0,5-----------------0,5 mol
nCH3COOC2H5=44\88=0,5 mol
=>mCH3COOH=0,5.60=30g
=>mC2H5OH=0,5.46=23 g
Câu 3:
CH3CH2OH viết gọn lại thành C2H5OH
\(n_{CH3COOH}=0,1\left(mol\right)\)
\(n_{C2H5OH}=\frac{6,9}{46}=0,15\left(mol\right)\)
\(n_{CH3COOC2H5}=0,075\left(mol\right)\)
\(\frac{n_{CH3COOH}}{1}< \frac{n_{C2H5OH}}{1}\left(0,1< 0,15\right)\)nên hiệu xuất được tính theo CH3COOH
\(PTHH:C_2H_5+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)
\(H=\frac{n_{CH3COOC2H5}.100}{n_{CH3COOH}}=\frac{0,075.100}{0,1}=75\%\)
Câu 4:
Ta có:
\(V_{C2H5OH}=\frac{8,4}{0,8}=10,5\left(l\right)\)
\(\Rightarrow m_{H2O}=300.1=300\left(g\right)\)
\(\Rightarrow C\%_{C2H5OH}=\frac{8,4}{8,4+300}.100\%=2,7\%\)
\(D_r=\frac{10,5}{10,5+300}.100\%=3,38^o\)
CH3COOH + Mg ---> CH3COOMg + 1/2H2
(mol) 0,026 0,026 0,013
a) nCH3COOMg = 2,13 : 83 = 0,026 mol
=> C\(_M\)CH3COOH = 0,026 : 0,02 = 1,3 M
b) V\(_{H2}\)= 0,013 . 22,4 = 0,2912(lit)
c) CH3COOH + NaOH ----> CH3COONa + H2O
a. CH3COOH + NaOH------> CH3COONa + H2O
Ta có :
n NaOH = 30.20%/100%=6g
=> n NaOH =6/40=0,15 mol
Theo PTHH : n CH3COOH= n NaOH=0,15 mol
=> C M CH3COOH =0,15/0,5=0,3M (500ml=0,5l)
b. 2CH3COOH + Na2CO3 ----------> CH3COONa + H2O + CO2
n Na2CO3 =0,5.0,2=0,1 mol
Ta có : n CH3COOH/2=0,15mol > n Na2CO3 =0,1mol
=> n CH3COOH dư, n Na2CO3 hết
Theo PTHH : n CO2 = n Na2CO3 = 0,1 mol
=> V CO2 ( ở đktc ) = 0,1.22,4=2,24l
Zn + 2CH3COOH => (CH3COO)2Zn + H2
nZn = m/M = 6.5/65 = 0.1 (mol)
Theo pt =>> nCH3COOH = 0.2 (mol)
==> mCH3COOH = n.M = 0.2 x 60 = 12 (g)
mdd = 12x100/12 = 100 (g)