Question

Answer 1

a, PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\)

b, Ta có: \(n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)

\(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\), ta được C₂H₅OH dư.

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

\(\Rightarrow n_{C_2H_5OH\left(dư\right)}=0,1\left(mol\right)\Rightarrow m_{C_2H_5OH\left(dư\right)}=0,1.46=4,6\left(g\right)\)

c, Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5}=0,2.88=17,6\left(g\right)\)

\(\Rightarrow H\%=\dfrac{14,08}{17,6}.100\%=80\%\)

d, Ta có: d_C2H5OH = 0,8 g/ml

⇒ V_C2H5OH = 13,8/0,8 = 17,25 (ml)

⇒ Thu được rượu có số độ là: \(\dfrac{17,25}{17,25+40,25}.100=30^o\)

Bạn tham khảo nhé!