Đáp án: $$80,91\%$$
Giải thích các bước giải:
$$CH3COOH+C_2H_5OH\rightleftharpoons CH_3COOC_2H_5+H_2O$$
$$n_{CH_3COOC_2H_5}=2$$ mol
$$\Rightarrow n_{CH_3COOH\text{pứ}}=n_{C_2H_5OH\text{pứ}}=n_{H_2}=2$$ mol
$$\Rightarrow n_{CH_3COOH\text{cb}}=n_{C_2H_5OH\text{cb}}=3-2=1$$ mol
Hằng số cân bằng:
$$K_c=\dfrac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}=\dfrac{n_{CH_3COOC_2H_5\text{cb}}.n_{H_2O\text{cb}}}{n_{CH_3COOH\text{cb}}.n_{C_2H_5OH\text{cb}}}=\dfrac{2.2}{1.1}=4$$
Khi $$n_{CH_3COOH}=3$$ mol; $$n_{C_2H_5OH}=5$$ mol:
$$n_{CH_3COOC_2H_5}=n_{H_2O}=x$$ mol
$$\Rightarrow n_{CH_3COOH\text{cb}}=3-x$$ mol; $$n_{C_2H_5OH\text{cb}}=5-x$$ mol
$$\Rightarrow \dfrac{x^2}{(3-x)(5-x)}=4$$ ($$x<3$$)
$$\Rightarrow x=2,4274$$
Theo lí thuyết, $$3$$ mol $$CH_3COOH$$ phản ứng hết
Vậy hiệu suất phản ứng là:
$$\dfrac{2,4274.100\%}{3}=80,91\%$$