If 8.76g of PbF2 preciptate is obtained when 4.50 g of KF is added to a solution which contains an excess of Pb(NO3)2, what is the precent yield?
2 KF(s) + Pb(NO3)2(aq) -> 2 KNO3(aq) + PbF2(s)
2006-07-12 21:48:56 · 3 answers · asked by nba_joker 1 in Science & Mathematics ➔ Mathematics
1 mole of KF has mass 58 g, so we have 4.50/58 = 0.0776 mole. Because the coefficient of KF is two, 0.0776/2 = 0.0388 mole of the reaction could take place.
1 mole of PbF2 has mass 245.2 g, so the yield is 8.76/245.2 = 0.0357 mole of the reaction.
The percentage is 0.0357/0.0388 = 92%.
2006-07-22 13:35:13 · answer #1 · answered by dutch_prof 4 · 0⤊ 0⤋
ask thios question in chemistry section not mathematics
2006-07-13 12:21:43 · answer #2 · answered by Theta40 7 · 0⤊ 0⤋
Now now, we can't be doing your homework for you...
2006-07-13 04:50:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋