\(n_{H_2SO_4}=0,3\times0,75=0,225\left(mol\right)\)
\(n_{HCl}=0,3\times1,5=0,45\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(0,45\) \(\leftarrow0,225\) \(\left(mol\right)\)
\(KOH+HCl\rightarrow KCl+H_2O\)
\(0,45\) \(\leftarrow0,45\) \(\left(mol\right)\)
Vậy \(n_{KOH}=0,45+0,45=0,9\left(mol\right)\)
\(\Rightarrow V_{KOH}=\dfrac{n_{KOH}}{C_{M_{KOH}}}=\dfrac{0,9}{1,5}=0,6\left(l\right)=600\left(ml\right)\)