According to the following reaction, what mass (in g) of PbF2 can form from 51.3 mL of 0.12 M NaF solution and 290.7 mL of 0.26 M Pb(NO3)2?

(Molar masses: PbF2 = 245.2 g/mol; NaF = 41.99 g/mol; Pb(NO3)2 = 331.2 g/mol) 2 NaF(aq) + Pb(NO3)2(aq) PbF2(s) + 2 NaNO3(aq)