Question

Sven perform this reaction with 15.0 grams of sodium sulfate and an excess of iron (III) phosphate to make iron (III) sulfate and sodium phosphate. In the actual experiment, 10.0 grams of sodium phosphate are experimentally made, what is the percent yield?

2FePO4+3Na2SO4-->Fe2(SO4)3+2Na3PO4

Answer 1

85.5%

Step-by-step explanation:

To get the experimental value, you need to convert 15.0 grams of Na2SO4 to grams of Na3PO4. You do this with stoichiometry.

Convert grams of Na2SO4 to moles with molar mass. Then convert to moles of Na3PO4 with the mole-to-mole ratio according to the balanced chemical equation. Then convert moles of Na3PO4 to grams with the molar mass.

15.0 g Na2SO4 x (1 mol/142.04 g) x (2 Na3PO4/3Na2SO4) x (163.94 g/1 mol) = 11.7 g Na3PO4

Percent Yield = (actual value/experimental value) x 100

Actual Value = 10.0 g

Experimental Value = 11.7 g

10.0g/11.7 g = 85.5%