$n_{CO_2}=n_{CaCO_3}=\dfrac{25,5}{100}=0,255(mol)$
$\Delta m_{dd}=m_{CaCO_3}-m_{H_2O}-m_{CO_2}$
$\Rightarrow m_{H_2O}=25,5-0,255.44-9,87=4,41g$
$\Rightarrow n_{H_2O}=0,245(mol)$
Bảo toàn khối lượng:
$m_{O_2}=0,255.44+0,245.18-4,03=11,6g$
$\Rightarrow n_{O_2}=0,3625(mol)$
Trong X có 6 nguyên tử O nên $n_{O(X)}=6n_X$
Bảo toàn O: $6n_X+2n_{O_2}=2n_{CO_2}+n_{H_2O}$
$\Rightarrow n_X=0,005(mol)$
Khi $m_X=8,06g$, $n_X=0,005.2=0,01(mol)$
$n_{NaOH}=0,01.3=0,03(mol)$
$n_{C_3H_5(OH)_3}=0,01(mol)$
Bảo toàn khối lượng:
$m=8,06+0,03.40-0,01.92=8,34g$
