$\%O=11.354\%$

$n_{CO_2}=n_C=\dfrac{76.641 \%}{12}m(mol)$

$n_{H_2O}=\dfrac{n_H}{2}=\dfrac{12.005 \%}{2}m(mol)$

$n_O=\dfrac{11.354\%}{16}(mol)$

Bảo toàn nguyên tố $O:$

$n_X=\dfrac{n_O}{6}=\dfrac{11.354\%}{96}m(mol)$

$0.1m(gam)$ $X$ phản ứng $0.05(mol)$ $H_2.$

`->`$m(gam)$ $X$ phản ứng $0.5(mol)$ $H_2.$

`=>`$n_X=\dfrac{n_{CO_2}-n_{H_2O}-n_{H_2}}{k-1}$

`<=>`$\dfrac{11.354\%}{96}m=\dfrac{\dfrac{76.641 \%}{12}m-\dfrac{12.005 \%}{2}m-0.5}{3-1}$

`<=>`$m \approx 338.505(gam)$

$n_X=\dfrac{11.354\%}{96} \times 338.505 \approx 0.4(mol)$

`=>`$n_{KOH}=3 \times 0.4=1.2(mol)$

$n_{C_3H_8O_3}=n_X=0.4(mol)$

Bảo toàn khối lượng:

$338.505+56 \times 1.2=a+92 \times 0.4$

`=>`$a=368.905$