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nKOH = 0,5.0,3 = 0,15 mol
CH3COOH + KOH → CH3COOK + H2O
0,15 0,15 0,15 mol
a) CM CH3COOH = 0,15/0,2 =0,75M
b) Thể tích của dung dịch thu được sau phản ứng: 500 ml
CM CH3COOK = 0,15/0,5 = 0,3M
c) Phản ứng lên men giấm
C2H5OH + O2 → CH3COOH + H2O
0,15 0,15
→ mC2H5OH = 0,15.46 = 6,9 gam
\(n_{KOH}=0,5\cdot0,3=0,15mol\)
\(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,15 0,15 0,15 0,15
a)\(C_{M_{CH_3COOH}}=\dfrac{0,15}{0,2}=0,75M\)
b)\(C_{M_{CH_3COOK}}=\dfrac{0,15}{0,2+0,3}=0,3M\)
a) \(n_{CH_3COOH}=\dfrac{120.20\%}{60}=0,4\left(mol\right)\)
\(n_{Na_2CO_3}=\dfrac{53.30\%}{106}=0,15\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\) => Na2CO3 hết, CH3COOH dư
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,15-------->0,3-------------->0,3------->0,15
=> \(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\)
b) mdd sau pư = 120 + 53 - 0,15.44 = 166,4 (g)
=> \(C\%=\dfrac{24,6}{166,4}.100\%=14,78\text{%}\)
\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ a,n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\Rightarrow n_{CO_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{50}{200}.0,2=0,05\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,05}{1}\Rightarrow Rượu.dư\\ \Rightarrow n_{este\left(LT\right)}=n_{axit}=0,05\left(mol\right)\\ \Rightarrow n_{este\left(TT\right)}=80\%.0,05=0,04\left(mol\right)\\ m_{CH_3COOC_2H_5}=88.0,04=3,52\left(g\right)\)
\(a,m_{Na_2CO_3}=\dfrac{500.20}{100}=100\left(g\right)\\ \rightarrow n_{Na_2CO_3}=\dfrac{100}{106}=\dfrac{50}{53}\left(mol\right)\)
PTHH: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2\uparrow+H_2O\)
\(\dfrac{50}{53}\)------->\(\dfrac{100}{53}\)--------------->\(\dfrac{100}{53}\)-------------->\(\dfrac{50}{53}\)
\(b,m_{axit}=\dfrac{100}{53}.60=\dfrac{6000}{53}\left(g\right)\\ c,m_{dd}=500+400-\dfrac{50}{53}.44=\dfrac{45500}{53}\left(g\right)\\ m_{CH_3COONa}=\dfrac{100}{53}.82=\dfrac{8200}{53}\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{\dfrac{8200}{23}}{\dfrac{45500}{23}}.100\%=18,02\%\)
\(a.n_{CO_2}=\dfrac{0,672}{22,4}=0,03mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
\(n_{CO_2}=n_{Ca\left(OH\right)_2}=n_{CaCO_3}=0,03mol\\ m_{CaCO_3}=0,03.100=3g\\ b.V_{ddCa\left(OH\right)_2}=\dfrac{0,03}{1,5}=0,02l\)
Rượu etylic \(C_2H_5OH\)
Axit axetic \(CH_3COOH\)
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+H_2O+CO_2\uparrow\)
0,2 0,1 0,2 0,1 0,1
\(\%m_{CH_3COOH}=\dfrac{0,2\cdot60}{39,6}\cdot100\%=30,3\%\)
\(\%m_{C_2H_5OH}=100\%-30,3\%=69,7\%\)
a)
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
0,2<----------0,1<-------------0,2<-------0,1
=> \(m_{CH_3COOH}=0,2.60=12\left(g\right)\)
\(\%m_{CH_3COOH}=\dfrac{12}{39,6}.100\%=30,3\%\)
\(\%m_{C_2H_5OH}=\dfrac{39,6-12}{39,6}.100\%=69,7\%\)
b) dd sau pư chứa \(\left\{{}\begin{matrix}CH_3COONa:0,2\left(mol\right)\\C_2H_5OH:\dfrac{39,6-12}{46}=0,6\left(mol\right)\end{matrix}\right.\)
\(V_{dd}=\dfrac{0,1}{2}=0,05\left(l\right)\)
=> \(\left\{{}\begin{matrix}C_{M\left(CH_3COONa\right)}=\dfrac{0,2}{0,05}=4M\\C_{M\left(C_2H_5OH\right)}=\dfrac{0,6}{0,05}=12M\end{matrix}\right.\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)
b, mZn = 0,15.65 = 9,75 (g)
c, CM (H2SO4) = 0,15/0,05 = 3 M
d, mZnSO4 = 0,15.161 = 24,15 (g)
Bạn tham khảo nhé!
nCO2 = 0,448 / 22,4 = 0,02 (mol)
2CH3COOH + CaCO3 -> (CH3COO)2Ca + CO2 + H2O
0,04 0,02 0,02
mCH3COOH = 0,04 * 60 = 2,4 (gam)
-> C%CH3COOH = 2,4 / 40 * 100% = 6%
mCaCO3 = 0,02 * 100 = 2 (gam)