Ta có: 40nNaOH + 56nKOH = 3,04 (1)
PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)
\(KOH+HCl\rightarrow KCl+H_2O\)
Theo PT: \(n_{NaCl}=n_{NaOH}\)
\(n_{KCl}=n_{KOH}\)
⇒ 58,5nNaOH + 74,5nKOH = 4,15 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,02\left(mol\right)\\n_{KOH}=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%m_{KOH}=\dfrac{0,04.56}{3,04}.100\%\approx73,68\%\)
→ Đáp án: C