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a, Ta có: \(n_{CH_3COOH}=\dfrac{9,6}{60}=0,16\left(mol\right)\)
PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1}{2}n_{CH_3COOH}=0,08\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,08.142=11,36\left(g\right)\)
b, PT: \(CH_3COOH+C_2H_5OH\underrightarrow{t^o,xt}CH_3COOC_2H_5+H_2O\)
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,16\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,16.88=14,08\left(g\right)\)
Mà: thực tế thu được 10,56 (g)
\(\Rightarrow H\%=\dfrac{10,56}{14,08}.100\%=75\%\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,1 0,2
a. \(V_{CH_3COOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)
b. \(CH_3COOH+C_2H_5OH⇌\left(H_2SO_{4đ},t^o\right)CH_3COOC_2H_5+H_2O\)
0,2 0,2
Với H% = 80
\(m_{CH_3COOC_2H_5}=\dfrac{0,2.88.80}{100}=14,08\left(g\right)\)
\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ a,n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\Rightarrow n_{CO_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{50}{200}.0,2=0,05\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,05}{1}\Rightarrow Rượu.dư\\ \Rightarrow n_{este\left(LT\right)}=n_{axit}=0,05\left(mol\right)\\ \Rightarrow n_{este\left(TT\right)}=80\%.0,05=0,04\left(mol\right)\\ m_{CH_3COOC_2H_5}=88.0,04=3,52\left(g\right)\)
a, \(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\)
\(\Rightarrow a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\)
b, \(m_{dd}=\dfrac{15}{2\%}=750\left(g\right)\)
\(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\\ NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\\a, n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\\ a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\\ b,m_{ddCH_3COOH}=\dfrac{15.100}{2}=750\left(g\right)\)
a, PT: \(CH_3COOH+C_2H_5OH\underrightarrow{_{H_2SO_{4\left(đ\right)}}}CH_3COOC_2H_5+H_2O\)
b, Ta có: \(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)
\(n_{C_2H_5OH}=\dfrac{27,6}{46}=0,6\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\), ta được C2H5OH.
Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,5\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,5.88=44\left(g\right)\)
Mà: m CH3COOC2H5 (TT) = 35,2 (g)
\(\Rightarrow H\%=\dfrac{35,2}{44}.100\%=80\%\)
Bạn tham khảo nhé!
\(n_{NaOH}=0,2.1,5=0,3\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,1---------->0,1
=> mNaOH = 0,1.40 = 4 (g)
=> \(C\%_{NaOH}=\dfrac{4}{80}.100\%=5\%\)
PTHH: CH3COOH + C2H5OH --H2SO4(đặc), to--> CH3COOC2H5 + H2O
0,3------------------------------------------------>0,3
=> meste = 0,3.88.80% = 21,12 (g)
CH3COOH+NaOH->CH3COONa+H2O
0,3--------------0,3-------0,3--------------0,3
n CH3COOH=0,3 mol
=>C% NaOH=\(\dfrac{0,3.40}{80}.100\)=15%
H=80%
=>m CH3COOC2H5=0,3.88.80%=21,12g