Answer
$V(HCl)=464.1ml$
Work Step by Step
The following is the neutralization reaction of hydrochloric acid by calcium hydroxide:
$2HCl + Ca(OH)_{2} \rightarrow CaCl_{2} + 2H_{2}O$
By considering stoichiometric coefficients, we can observe that the amount of hydrochloric acid required is twice the amount of calcium hydroxide. Hence,
$n(HCl)=2n(Ca(OH)_{2})=2\times \frac{m(Ca(OH)_{2})}{M(Ca(OH)_{2})}=2\times \frac{1.98g}{74\frac{g}{mol}}=0.0535mol$
We can obtain the volume of $HCl$ solution knowing its concentration:
$V(HCl)=\frac{n(HCl)}{c(HCl)}=\frac{0.0535mol}{0.1153\frac{mol}{dm^3}}=0.4641dm^3=464.1ml$
