Question

Answer 1

a, \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=0,2\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH\left(LT\right)}=0,2.60=12\left(g\right)\)

\(\Rightarrow H=\dfrac{9,5}{12}.100\%\approx79,167\%\)

b, Ta có: n_{C2H5OH còn lại} = 0,2 - 0,2.79,167% ≃ 0,042 (mol)

\(n_{CH_3COOH}=\dfrac{9,5}{60}=\dfrac{19}{120}\left(mol\right)\)

PT: \(C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄ đặc, t^o)

Xét tỉ lệ: \(\dfrac{0,042}{1}< \dfrac{\dfrac{19}{120}}{1}\), ta được CH₃COOH dư.

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{C_2H_5OH}=0,042\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,042.88=3,696\left(g\right)\)

Mà: H = 75%

\(\Rightarrow m_{CH_3COOH\left(TT\right)}=3,696.75\%=2,772\left(g\right)\)