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\(n_{Na2CO3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH3OONa+CO_2+H_2O\)
0,2 0,1 0,2 0,1
a) \(V_{CO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
\(C_{MCH3COOH}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
b) \(C_{MCH3COONa}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Chúc bạn học tốt
a, \(n_{K_2CO_3}=\dfrac{2,76}{138}=0,02\left(mol\right)\)
PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)
Theo PT: \(n_{CH_3COOH}=2n_{K_2CO_3}=0,04\left(mol\right)\)
\(\Rightarrow C\%_{CH_3COOH}=\dfrac{0,04.60}{50}.100\%=4,8\%\)
b, \(C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có: \(n_{CH_3COONa}=\dfrac{9,84}{82}=0,12\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=n_{NaOH}=n_{CH_3COONa}=0,12\left(mol\right)\)
\(\Rightarrow V_{ddCH_3COOH}=\dfrac{0,12}{0,5}=0,24\left(l\right)\)
\(m_{NaOH}=0,12.40=4,8\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{4,8}{20\%}=24\left(g\right)\)
\(a.Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{Zn}=\dfrac{4,55}{65}=0,07\left(mol\right)\\ n_{H_2}=n_{Zn}=0,07\left(mol\right)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568\left(l\right)\\ b.n_{HCl}=2n_{Zn}=0,14\left(mol\right)\\ \Rightarrow CM_{HCl}=\dfrac{0,14}{0,2}=0,7M\\ c.n_{ZnCl_2}=n_{Zn}=0,07\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52\left(g\right)\)
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\\ n_{CuO}=\dfrac{37}{80}\left(mol\right)=n_{\left(CH_3COO\right)_2Cu}\\ n_{CH_3COOH}=2.\dfrac{37}{80}=\dfrac{37}{40}\left(mol\right)\\ V_{ddCH_3COOH}=\dfrac{\dfrac{37}{40}}{2}=\dfrac{37}{80}\left(l\right)\\ C_{Mdd\left(CH_3COO\right)_2Cu}=\dfrac{\dfrac{37}{80}}{\dfrac{37}{80}}=1\left(M\right)\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,n_{CH_3COOH}=2.n_{H_2}=2.0,3=0,6\left(mol\right)\\ m_{ddCH_3COOH}=\dfrac{0,6.60.100}{20}=180\left(g\right)\\ b,n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ C\%_{dd\left(CH_3COO\right)_2Zn}=\dfrac{0,3.183}{180+0,3.65-0,3.2}.100\approx27,602\%\)
`2CH_3COOH+Zn->(CH_3COO)_2Zn+H_2`
0,2----------------------------0,1 mol
`n_((CH_3COO)_2Zn)=(18,3)/183=0,1 mol`
`->CM(CH_3COOH)=(0,2)/(0,5)=0,4 M`
`#YBTran:3`