Question

Consider the following balanced equation:3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)If 24.2 moles of Na3PO4(aq) reacts with an excess of Ca(NO3)2(aq) to produce 50.6 moles of NaNO3(aq), what is the percent yield of the reaction?

Answer 1

The percent yield of the reaction is 69.8%.

Step-by-step explanation:

The percent yield of a reaction measures how effectively the reactants form the desired product. It is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. In this case, the actual yield is given as 50.6 moles of NaNO3(aq). The theoretical yield can be calculated from the balanced equation by determining the mole ratio between Na3PO4(aq) and NaNO3(aq). Since the ratio is 2:6, the theoretical yield of NaNO3(aq) is (6/2) * 24.2 moles = 72.6 moles. Therefore, the percent yield is (50.6 / 72.6) * 100 = 69.8%.

Answer 2

69.7% is percent yield

Step-by-step explanation:

Based on the reaction:

3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

2 moles of Na3PO4 react producing 6 moles of NaNO3.

As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:

24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =

72.6 moles of NaNO3

As there are produced 50.6 moles of NaNO3, percent yield is:

50.6 moles NaNO3 / 72.6 moles NaNO3 =

69.7% is percent yield