\(C_2H_5COOK\)+KOH→\(C_2H_5COOK\)+\(H_2O\)
\(n_{C_2H_5COOK}=\dfrac{16,8}{112}\)=0,15(mol)
nC2H5COOH=nC2H5COOK=0,15(mol)
VddC2H5COOH=\(\dfrac{0,15}{1}\)=0,15(l)=150(ml)
⇒V=150(ml)
+Koh→+
=0,15(mol)
nC2H5COOH=nC2H5COOK=0,15 mol
VddC2H5COOH==0,15(l)=150(ml)
= V=150(ml)