#2) Consider the preciditation reaction of a solution containing 3.50 grams of sodium phosphate mixed with a solution containing 6.40 grams of barium nitrate. How many grams of Barium phosphate can form?
Ba(NO3 )2 + Na3 PO4 yields Ba3 (PO 4 )2 + NaNO3
#2b) If the actural yield of the previous problem was 4.70 grams of barium phosphate. What is the % yield?
i got 4.91g Ba3(PO4)2 as the limiting reagent for number 2. so how do i do 2b??
I checked it, too, and 4.01 Ba3(PO4)2 is correct BUT Ba3(PO4)2 is not the limiting reagent. The limiting reagent is the Ba(NO3)2. Ba3(PO4)2 is the product.
How to do the percent?
The 4.91 g Ba3PO4)2 is the theoretical yield. The actual yield is 4.70 g.
%yield = (actual yield)/(theoretical yield)*100.
To find the % yield in this problem, you need to compare the actual yield to the theoretical yield. The theoretical yield is the calculated amount of barium phosphate that should be formed based on stoichiometry and the limiting reagent.
Step 1: Calculate the theoretical yield
In question #2, you calculated that the limiting reagent was barium nitrate and found that the amount of barium phosphate formed was 4.91 grams.
Step 2: Calculate the % yield
The % yield is calculated using the formula:
% Yield = (Actual Yield / Theoretical Yield) x 100
In this problem, the actual yield is given as 4.70 grams. Plug in the values into the formula:
% Yield = (4.70 g / 4.91 g) x 100
% Yield = 95.93%
So, the % yield in this reaction is approximately 95.93%.