Câu 11:
Đặt $n_{CO_2}=y; n_{O_2}=x(mol)$
BTKL: $25,74+32x=44y+1,53.18$
$\to 32x-44y=1,8$
$n_{COO}=\dfrac{2y+1,53-2x}{2}=-x+y+0,765(mol)$
$\to n_X=\dfrac{n_{COO}}{3}=\dfrac{-x+y+0,765}{3}(mol)$
$n_{CO_2}-n_{H_2O}=(k-1)n_X=n_{Br_2}+2n_X$
$\to y-1,53=0,06+2.\dfrac{-x+y+0,765}{3}$
Giải hệ: $x=2,325; y=1,65$
$\to n_{COO}=0,09(mol)=n_{NaOH}$
$\to n_{C_3H_8O_3}=0,03(mol)$
BTKL $\to m=26,58g$
$\to B$
