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Mg + 2CH3COOH => (CH3COO)2Mg + H2
n muối = m/M = 1.42/142 = 0.01 (mol)
Theo phương trình ==> nCH3COOH = 0.02 (mol). 50ml = 0.05 (l)
CM dd CH3COOH = n/V = 0.02/0.05 = 0.4 M
CH3COOH + NaOH => CH3COONa + H2O
50ml---> 0.02 mol; 100ml => 0.04 mol
==> nNaOH = 0.04 (mol)
V dd NaOH = n/Cm = 0.04/0.5 = 0.08 l = 80ml
a)\(KOH+CH3COOH-->H2O+CH3COOK\)
\(m_{CH3COOH}=\frac{100.12}{100}=12\left(g\right)\)
\(n_{CH3COOH}=\frac{12}{60}=0,2\left(mol\right)\)
\(n_{KOH}=n_{CH3COOH}=0,2\left(mol\right)\)
\(m_{KOH}=0,2.56=11,2\left(g\right)\)
\(m_{ddKOH}=\frac{11,2.100}{10}=112\left(g\right)\)
b)\(mdd\) sau pư =\(100+112=212\left(g\right)\)
\(n_{CH3COOK}=n_{CH3COOH}=0,2\left(mol\right)\)
\(m_{CH3COOK}=0,2.98=19,6\left(g\right)\)
\(C\%_{CH3COOK}=\frac{19,6}{212}.100\%=9,25\%\)
c)
2CH3COOH + Mg => (CH3COO)2Mg + H2
nZn = m/M = 13/65 = 0.2 (mol)
Đặt số mol lên phương trình ta được:
nCH3COOH = 0.4 (mol)
nH2 = n(CH3COO)2Mg = 0.2 (mol)
VH2 = 22.4 x 0.2 = 4.48 (l)
mCH3COOH = n.M = 0.4 x 60 = 24 (g)
C%dd CH3COOH = 24 x 100/200 = 12 (g)
m(CH3COO)2Mg = 142x 0.2 = 28.4 (g)
mdd sau pứ = 200 + 13 - 0.2 x 2 = 212.6 (g)
C% dd (CH3COO)2Mg = 28.4 x 100/212.6 = 13.36 %
mdd sau pứ = mdd CH3COOH + mZn - mH2
Mà mH2 = n.M = 0.2 x 2 đó bn
1/ Đổi: 15.2ml = 0.0152l; 10ml = 0.01l
CH3COOH + NaOH => CH3COONa + H2O
nNaOH = CM.V = 0.0152x0.2 = 19/6250 (mol)
Theo phương trình ==> nCH3COOH = 19/6250 (mol)
CM ddCH3COOH = n/V = 0.304M
2/ mC = 2.4 (g), mH = 0.6 (g)
=> mO = 4.6 - (2.4 + 0.6) = 1.6 (g)
CT: CxHyOz
Theo đề bài ta có: 12x/2.4 = y/0.6 = 16z/1.6 = 46/4.6
=> x = 2, y = 6, z =1
CT: C2H6O hay C2H5OH
CH3COOH + NaHCO3 \(\rightarrow\) CH3COONa + H2O + CO2\(\uparrow\) (1)
a) Ta có: \(m_{NaHCO_3}=\dfrac{20\cdot500}{100}=100\left(g\right)\)
\(\Rightarrow n_{NaHCO_3}=\dfrac{100}{84}\approx1,2\left(mol\right)\)
Theo phương trình (1): \(n_{CH_3COOH}=n_{NaHCO_3}=1,2\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=1,2\cdot60=72\left(g\right)\)
Do đó: \(C\%_{CH_3COOH}=\dfrac{72}{300}\cdot100\%=24\%\)
a)
\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,15<---------0,3<------------------------0,15
=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)
b)
\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)
c)
PTHH: 2CH3COOH + Ba(OH)2 --> (CH3COO)2Ba + 2H2O
0,3--------->0,15
=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)
ADCT: nCaCO3=m/M=50/100=0,5(mol)
a,PTHH: CaCO3+2CH3COOH-->(CH3COO)2Ca+CO2+H2O
b, Theo pt: 1 mol CaCO3: 2 mol CH3COOH: 1 mol (CH3COOH)2Ca: 1 mol CO2
Theo đb: 0,5 mol CaCO3: x mol CH3COOH: y mol (CH3COOH)2Ca: z mol CO2
-->x=1 mol
-->y=0,5 mol
-->z=0,5 mol -->mCO2=n.M=0,5.44=22(g)
ADCT: mCH3COOH=n.M=1.60=60(g)
ADCT: C%CH3COOH= (mct/mdd).100%=(60/200).100=30(g)
c, ADCT: m(CH3COOH)2Ca=n.M=0,5.120=60(g)
-->mdd(CH3COOH)2Ca sau p/ứ=(50+200)-22=228(g)
ADCT: C%(CH3COOH)2Ca=(mct/mdd).100%=(60/228).100~26,31(%)
Vậy b, C%CH3COOH=30%
c, C%(CH3COOH)2Ca~26,31 %
áp dụng sơ đồ đường chéo ta có:
50 0 40 40 10
\(\Rightarrow\) \(\frac{m1}{m2}\)= \(\frac{40}{10}\)= \(\frac{4}{1}\)
\(\Rightarrow\) \(\frac{200}{m2}\)= \(\frac{4}{1}\)
\(\Rightarrow\) m2= \(\frac{200}{4}\)= 50( g)
vậy cần thêm 50g nước vào 200g dd CH3COOH 50% để thu được dd CH3COOH 40%