Ta có: \(m_{HCl}=50.14,6\%=7,3\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{0,2}{0,5}=0,4\left(l\right)=400\left(ml\right)\)
KOH + HCl → KCl + H2O
\(0,146\) \(0,146\) \(0,146\)
\(V_{KOH}=\dfrac{n}{CM}=\dfrac{0,146}{0,5}=0,292\left(l\right)=292\left(ml\right)\)