What is the percent yield if 0.3 mol ba(no3)2 and 0.25 mol na3po4 react to produce 0.095 mol ba3(po4)2?

The chemical reaction is expressed as:

3Ba(NO3)2 + 2Na3PO4 = Ba3(PO4)2 + 6NaNO3

To determine the percent yield, we need to determine the theoretical yield of the reaction from the given amounts of the reactants. We do as follows:

0.3 mol 3Ba(NO3)2 ( 2 mol Na3PO4 / 3 mol Ba(NO3)2) = 0.2 mol Na3PO4

Therefore, the limiting reactant would be Ba(NO3)2 since it is consumed completely in the reaction.

Theoretical yield = 0.3 mol 3Ba(NO3)2 ( 1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.1 mol Ba3(PO4)2

Percent yield = actual yield / theoretical yield = 0.095 mol Ba3(PO4)2 / 0.1 mol Ba3(PO4)2 x 100 = 95%