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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?

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jufsanabriasa

Answer:

69.7% is percent yield

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

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