Answer
$n(HCl)= 7.6mmol$
$V(HCl)=12.54ml$
Work Step by Step
The following reaction occurs:
$NaOH + HCl \rightarrow NaCl + H_{2}O$
Considering stoichiometric coefficients, we can conclude that
$n(HCl)=n(NaOH)=c(NaOH)V(NaOH)=0.298\frac{mol}{dm^3}\times 0.0255dm^3=0.007599mol\approx 7.6mmol$
$V(HCl)=\frac{n(HCl)}{c(HCl)}=\frac{0.007599mol}{0.606\frac{mol}{dm^3}}=0.01254dm^3=12.54ml$
