nglyxeron = nx= nglyxeron = x
→ nKOH =3nx → nKOH = 3x → mKOH =168x → mddKOH =168x/28.100= 600x → mH2O = 600x-168x=432x
Theo đề: mH2O + mglyxeron = 26,2 → x=0,05 mol
432 92
Ta có: nKOH = 0,15 mol → 42,38 + 0,15.56 = mRCOOK + 0,05.92 → mRCOOK = 46,18 gam
Phương trình đốt cháy muối: \( \begin{align}
& 2{{C}_{x}}{{H}_{y}}{{O}_{2}}K+\left( \left. 2x+\frac{y}{2}-1 \right) \right.{{O}_{2}}\xrightarrow{{}}(2x-1)C{{O}_{2}}\,\,+\,\,\,y{{H}_{2}}O+{{K}_{2}}C{{O}_{3}} \\
& \,\,\,\,\,\,\,\,0,15\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2x-1).0,075\,\,\,\,\,0,075y \\
\end{align}\)
Ta có: lk trong -C- C- = nBr2 /nx → nBr2= 0,21
