For the reaction BaCl2 (aq) + ZnSO4 (aq) BaSO4 precipitates out.Calculate the percent yield (%)?

Question

when 2.050 L of 0.0550 M BaCl2 (aq) and 2.020 L of 0.0429 M ZnSO4 (aq) are mixed.

Answer 1

OK...This one's going to take a bit to talk you through.

First you have to determine which of your two reactants is the limiting reactant. That's the one that will give you the smallest possible amount of the product.. Hopefully, you can see that barium ions and sulfate ions react in a 1:1 ratio. If you write out the balanced equation, you'll see that.

So, calculate the number of moles of each reactant that you are starting with from each solution's volume and molarity. The reactant that you have the smallest number of moles of is your limiting reactant (this isn't ALWAYS the case, but is here since they react in a 1:1 ratio).

From the number of moles of that reactant, you know (again, from the balanced equation) that you'll produce that many moles of barium sulfate. You can then use the molar mass of BaSO4 to calculate the theoretical yield of barium sulfate.

(You're almost finished!!!).

Now, since you know the theoretically possible yield of BaSO4, to get the percent yield, just divide the mass that you got (13.084 grams) by the theoretical yield, and multiply that by 100 to make it a percent.

Answer 2

locate b/w both recommendations, the only which is in extra. (use n=CV) Use the answer that isn't in extra (as it truly is the most the recommendations will react) locate the mass of this answer ( n=Mass/Mr) and then to locate the yield purely divide the useful result with the aid of the "theory "result