Đáp án:

$\rm a)$

Gọi: $\begin{cases}\rm n_{C_nH_{2n+1}COOH}=a\ (mol)\\\rm n_{C_mH_{2m+1}OH}=b\ (mol)\end{cases}$

$\rm n_{CO_2}=\dfrac{15,68}{22,4}=0,7\ (mol)\\n_{H_2O}=\dfrac{15,3}{18}=0,85\ (mol)$

$\rm n_{C_mH_{2m+1}OH}=b=0,85-0,7=0,15\ (mol)$

Tăng giảm khối lượng:

$\rm n_{O_2}=\dfrac{0,7×44+15,3-18,9}{32}=0,85\ (mol)$

Bảo toàn nguyên tố O:

$\rm 2n_{C_nH_{2n+1}COOH}+n_{C_mH_{2m+1}OH}=2n_{CO_2}+n_{H_2O}\\⇔2a+0,15+2×0,85=0,7×2+0,85\\⇔a=0,2\ (mol)$

Bảo toàn nguyên tố C:

$\rm 0,2×(n+1)+0,15×m=0,7\\⇔2n+1,5m=5$

Thỏa mãn: $\rm n=1, m=2$

Vậy CTPT của:

$\rm R: CH_3COOH\\T:C_2H_5OH$

$\rm b)$

$\rm n_{H_2}=\dfrac{2,8}{22,4}=0,125\ (mol)$

Phản ứng xảy ra (mol):

$\rm CH_3COOH+C_2H_5OH\overset{H_2SO_4\ (dac),t^o}\leftrightarrows CH_3COOC_2H_5+H_2O\\CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2\\\hspace{,4cm} 0,2-x\hspace{2,35cm}\to\hspace{2cm}\dfrac{0,2-x}{2}\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2\\\hspace{,05cm}0,15-x\hspace{2,3cm}\to\hspace{1,cm}\dfrac{0,15-x}{2}\\H_2O+Na\to NaOH+\dfrac{1}{2}H_2\\\hspace{,3cm}x\hspace{2cm}\to\hspace{1,5cm}\dfrac{1}{2}x$

Theo phương trình:

$\rm n_{H_2}=\dfrac{0,2-x}{2}+\dfrac{0,15-x}{2}+\dfrac{x}{2}=0,125\ (mol)$

$\rm\to x=0,1\ (mol)$

$\rm\to H=\dfrac{n_{C_2H_5OH\ (phan\ ung)}}{n_{C_2H_5OH}}=\dfrac{0,1}{0,15}×100\%=66,67\%$