In one experiment, 22.9841 grams of 75.25% pure Pb(NO3)2 (pure Pb(NO3)2 has a molar mass of 331.2 grams) was mixed with 51.2354 grams of 81.21% pure CsI (pure CsI has a molar mass of 259.80992 grams):
2CsI(aq) + Pb(NO3)2(aq) → 2CsNO3(aq) + PbI2(s)
After the reaction, 10.4025 grams of PbI2(s) (molar mass 461.0 grams) were collected. What is the percent yield of the reaction?
actual yield/theoretical yield x 100% = percent yield
actual yield = 10.4025 g
The Attempt at a Solution
The equation is balanced and I solved to find that Pb(NO3)2 is the limiting reagent but, when I convert it from moles into grams to do the percent yield, i get a number over 100%.
This is my attempt…
22.9841 g Pb(NO3)2/75.25% = 0.3054 g Pb(NO3)2 x 1 mol/331.2 g x 1 mol PbI2/1 mol Pb(NO3)2
= 9.222 x 10-4 PbI2 (limiting reagent)
51.2354 g CsI/81.21% = 0.6309 g CsI x 1 mol/259.80992 g x 1 mol PbI2/2 mol CsI
= 1.214 x 10-3 mol PbI2
9.222 x 10-4 mol PbI2 x 461.0 g PbI2/1 mol PbI2 = 0.4251 g PbI2
10.4025 g/0.4251 g x 100% = 244% ?!?!?!
Please help :( I'm not sure where I went wrong...