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?

Question

Chemistry

Xiomara Stafford

14 October, 06:16

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?

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Sophie Reed · Answer 1

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%