1. - Thí nghiệm 1:
2CH3COOH + 2Na \(\rightarrow\) 2CH3COONa + H2 (1)
2C2H5OH + 2Na \(\rightarrow\) 2C2H5ONa + H2 (2)
- Thí nghiệm 2: Chỉ có CH3COOH phản ứng, C2H5OH không phản ứng
CH3COOH + NaOH \(\rightarrow\) CH3COONa + H2O (3)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\)(mol)
\(n_{NaOH}=1\cdot0,2=0,2\)(mol)
Theo phương trình (3): \(n_{CH_3COOH}=n_{NaOH}=0,2\)(mol)
\(\Rightarrow m_{CH_3COOH}=0,2\cdot60=12\)(g)
Theo phương trình (1): \(n_{H_2\left(1\right)}=\dfrac{1}{2}\cdot n_{CH_3COOH}=\dfrac{1}{2}\cdot0,2=0.1\)(mol)
Mà \(n_{H_2}=n_{H_2\left(1\right)}+n_{H_2\left(2\right)}\)
\(\Rightarrow0,15=0,1+n_{H_2\left(2\right)}\)
\(\Rightarrow n_{H_2\left(2\right)}=0,15-0,1=0,05\)(mol)
Theo phương trình (2): \(n_{C_2H_5OH}=2n_{H_2\left(2\right)}=2\cdot0,05=0,1\)(mol)
\(\Rightarrow m_{C_2H_5OH}=46\cdot0,1=4,6\)(g)
Do đó: \(m_a=m_{CH_3COOH}+m_{C_2H_5OH}=12+4,6=16,6\)(g)
2. Ta có: \(\%m_{C_2H_5OH}=\dfrac{4,6}{16,6}\cdot100\%=27,71\%\)
\(\%m_{CH_3COOH}=\dfrac{12}{16,6}\cdot100\%=72,29\%\)