1. The problem statement, all variables and given/known data Consider: 2 mol Reactant A (aq) + 3 mol Reactant B (aq) = 6 mol Product A (aq) + 1 mol Product B (s) You have 0.00075 L of a 0.0514M solution of Reactant A. In the lab, you combine it with 0.042 L of 0.0753M of Reactant B. You get a solid product and an aqueous product (a precipitation reaction!). Please, find the limiting reagent. Then consider this: if you recover 0.0569 g of Product B, then what is your percent yield? Pretend Product B's weight is 310.18g/mol. 2. Relevant equations N/a 3. The attempt at a solution Finding the Limiting Reagent: Considering Reactant A [(7.5*10^-4 L Reactant A/x moles Reactant A] = [(1 L Reactant A)/0.054 moles Reactant A]= 3.855*10^-5 moles reactant A used [3.855*10^-5 moles reactant A/x moles Product B]=[2 moles Reactant A/1 mole Product B]= 1.9275*10^-5 moles product B made Considering Reactant B [0.042L Reactant B/x moles Reactant B]=[1 L Reactant B/0.0753 moles Reactant B]= 0.0032 moles Reactant B used [0.0032 moles Reactant B/ x moles Product B]=[3 moles Reactant B/1 mole Product B]= 0.0011 moles Product B made :.Limiting Reagent is Reactant A. Yields: Theoretical yield: 1.9275*10^-5 moles product B *(310.18g/mol)= 0.006 g Product B Percent Yield: (0.0569g/0.006g)*100 = 948.3% Well. That doesn't seem right, but I can't find where I went wrong.