What mass of precipitate can be produced when 120.0 mL of .130 M barium chloride and 160.0 mL of 0.140 M iron(Ill) sulfate are mixed?
Sulfate ion is in excess, and thus
We're asked to find the theoretical yield of the precipitate of a reaction, given the reactants and their amounts.
We'll first write a balanced chemical equation for this reaction:
#3"BaCl"_2(aq) + "Fe"_2"(SO"_4")"_3(aq) rarr 3"BaSO"_4(s) + 2"FeCl"_3(aq)#
What we need to do is find the limiting reactant, by first finding the moles of each reactant present (via the molarity equation) and then dividing the mole value by their respective coefficients; whichever reactant has the lower value is limiting.
Divide by respective coefficients to find limiting reactant:
Barium chloride is thus the limiting reactant.
Now we use the mole value of
Finally, we use the molar mass of