HF is prepared by reacting $\mathrm{CaF}_{2}$ with $\mathrm{H}_{2} \mathrm{SO}_{4}:$
$$
\mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(\ell) \rightarrow 2 \mathrm{HF}(g)+\mathrm{CaSO}_{4}(s)
$$
HF can be electrolyzed, in turn, when dissolved in molten KF to produce fluorine gas:
$$
2 \mathrm{HF}(\ell) \rightarrow \mathrm{F}_{2}(g)+\mathrm{H}_{2}(g)
$$
Fluorine is extremely reactive, so it is typically sold as a $5 \%$ mixture by volume in an inert gas such as helium. How much $\mathrm{CaF}_{2}$ is required to produce $500.0 \mathrm{L}$ of $5 \% \mathrm{F}_{2}$ in helium? Assume the density of $\mathrm{F}_{2}$ gas is $1.70 \mathrm{g} / \mathrm{L}$