Answer:
The % yield is 95.0 %
Explanation:
Step 1: Data given
Number of moles Ba(NO3)2 = 0.3 moles
Number of moles Na3PO4 = 0.25 moles
Number of Ba3(PO4)2 = 0.095 moles
Step 2: The balanced equation
3 Ba(NO3)2 + 2 Na3PO4 → 6 NaNO3 + Ba3(PO4)2
Step 3: Calculate the limiting reactant
For 3 moles Ba(NO3)2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3(PO4)2
Ba(NO3)2 is the limiting reactant. It will compeltely be consumed completely (0.3 moles). Na3PO4 is in excess. There will react 0.20 moles.
There will remain 0.25-0.20= 0.05 moles
Step 4: Calculate moles Ba3(PO4)2
For 3 moles Ba(NO3)2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3(PO4)2
For 0.3 moles Ba(NO3)2 we'll have 0.3/3 = 0.1 mol Ba3(PO4)2
Step 5: Calculate percent yield
% yield = (actual yield / theoretical yield) * 100%
% yield = (0.095 / 0.1) *100 %
% yield = 95.0 %
The % yield is 95.0 %
