Pb(NO3)2 (aq) + K2CrO4 (aq) ======> PbCrO4(s) + 2 KNO3
(aq)
1 mole of Pb(NO3)2 consume 1 mole of K2CrO4
number of mole of K2CrO4 is = molarity x volume in L.
                                                      Â
= 0.5056 x 0.025
                                                       Â
= 0.1264 mole.
so,0.1264 mole of lead (II) nitrate solution will needed to
consue all K2CrO4.
volume of Pb(NO3)2 = no.of mole/ molarity
                                   Â
=0.1264 / 0.2456 =0.05146 L
froduct is formed = 1 mole of PbCrO4 and 2 mole of KNO3
         mass of
product = 1x 323.2 + 2x 101.1
                                      Â
= 525.4 g
