Question

Use the References to access Important values II needed for this question. How many grams of PbBr2 will precipitate when excess CrBrz solution is added to 57.0 mL of 0.660 M Pb(NO3)2 solution? 3Pb(NO3)2(aq) + 2CrBr3(aq) —3PbBr2(s) + 2Cr(NO3)3(aq) Submit Answer Retry Entire Group 2 more group attempts remaining

Answer 1

The Given Reaction is: 3Pb(NO3), (aq) + 267B (aq) → 3PbBr2 (8) + 2 Cr(NO3)2 (ay) from the Stochiometric coefficient of this equation, we can say that 3 mole of Pb(NO3)2 Produce 3 moles of PbBr2 = 1 mole of Pb(NO3), Produce 1 mole of Pb Brz. So, in the 1st step, calculate the moles of Pb(NO3), reacting; Volume of Pb(NO3), 57.0 ml Molanity of Pb(NO3), 2 0.660 M. Molarity Number of moles Volenne of Pb(NO3), in L Molarity Number of moles 0.057L 0.660 Ma Number of moles of Pb(NO3)2 0.0571 No. of moles of Paina), 2 (0.660M) (0.0576) 0.0 376 moles. If I mole of Pb(NO3), Produce 1 moles of PbBr2 + 0.0376 mole of Pb(NO3) Produce 0.0376 mole of Pb Bra | weight in grau of Pb Borg = ( No. of moles / Molecular weight )) 4 Pb Bry) 1 PbBr ) Molecular weight of AB = 367.01 gimol. .. weight of Pbbre (in graws] = (0.0376 mole) (367, 01 gimde/ weight of Pb Brz (9) = | 13.8069 g Answer