NaOH+HCl----->NaCl+H2O
N\(_{HC_{ }l}=2.0,3=0,6\left(mol\right)\)
Theo pthh
n\(_{NaOH}=n_{HCl}=0,6\left(mol\right)\)
V\(_{NaOH}=\frac{0,6}{2}=0,3l=300ml\)
\(nHCl=0,6\left(mol\right)\)
\(PTHH:NaOH+HCl\rightarrow NaCl+H2O\)
\(\Rightarrow nNaOH=0,5\left(mol\right)\)
\(\Rightarrow VNaOH=0,3l=300ml\)