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$n_{NaOH} = \dfrac{50.10\%}{40} = 0,125(mol)$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
Theo PTHH :
$n_{CH_3COOH} = n_{CH_3COONa} = n_{NaOH} = 0,125(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,125.60}{8\%} = 93,75(gam)$
$m_{dd\ sau\ pư} = m_{dd\ CH_3COOH} + m_{dd\ NaOH} = 143,75(gam)$
$C\%_{CH_3COONa} = \dfrac{0,125.82}{143,75}.100\% = 7,13\%$
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,n_{CH_3COOH}=2.n_{H_2}=2.0,3=0,6\left(mol\right)\\ m_{ddCH_3COOH}=\dfrac{0,6.60.100}{20}=180\left(g\right)\\ b,n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ C\%_{dd\left(CH_3COO\right)_2Zn}=\dfrac{0,3.183}{180+0,3.65-0,3.2}.100\approx27,602\%\)
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: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)
Fe+2CH3COOH->(CH3COO)2Fe+H2
33\112-33\56---------------33\112
n Fe=\(\dfrac{33}{112}\) mol
=>m (CH3COO)2Fe=\(\dfrac{33}{112}\).174=51,267g
=>VCH3COOH=\(\dfrac{\dfrac{33}{56}}{3}=0,196l\)
\(n_{Fe}=\dfrac{16,5}{56}=0,29mol\)
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
0,29 0,58 0,29 0,29 ( mol )
\(V_{H_2}=0,29.22,4=6,496l\)
\(m_{\left(CH_3COO\right)_2Fe}=0,29.174=50,46g\)
\(C_{M_{CH_3COOH}}=\dfrac{0,29}{0,3}=0,96M\)
Ta có: \(n_{SO_3}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: SO3 + H2O ---> H2SO4
Theo PT: \(n_{H_2SO_4}=n_{SO_3}=0,2\left(mol\right)\)
Đổi 250ml = 0,25 lít
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,25}=0,8M\)
Chọn B
\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\\ n_{CuO}=\dfrac{37}{80}\left(mol\right)=n_{\left(CH_3COO\right)_2Cu}\\ n_{CH_3COOH}=2.\dfrac{37}{80}=\dfrac{37}{40}\left(mol\right)\\ V_{ddCH_3COOH}=\dfrac{\dfrac{37}{40}}{2}=\dfrac{37}{80}\left(l\right)\\ C_{Mdd\left(CH_3COO\right)_2Cu}=\dfrac{\dfrac{37}{80}}{\dfrac{37}{80}}=1\left(M\right)\)