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All final answers MUST be clearly written in the boxes provided. All problem set-up and work MUST be written on this document supporting your final answer to receive credit. No additional sheets attached. Problems with insufficient work will receive zero credit even if the correct answer is reported. 1. Calculate the percent yield for the reaction described if 125.62 grams of magnesium phosphate is produced from combining 175.0 grams of solid lithium phosphate with 500.0 mL of 6.00 M aqueous magnesium sulfate. (0.5 pt) % yield-

Explanation / Answer

Balanced equation is,

4 Li3PO4 + 6 MgSO4 ---------------> 2 Mg3(PO4)2 + 6 Li2SO4

Mass of lithium phosphate = 175.0 g.

Molar mass of lithium phosphate = 115.8 g./mol

Moles of lithium phosphate = mass / molar mass = 175.0 / 115.8 = 1.511 mol

Moles og magnesium sulphate = molarity * volume 1000 = 6.00 * 500.0 / 1000 = 3.00 mol

From the balanced equation,

4 mol of lithium phosphate needs 6 mol of magnesium sulphate

then, 1.511 mol of lithium phosphate needs 1.511 * 6 / 4 = 2.27 mol of magnesium sulphate

But we have 3.00 mol of magnesium sulphate.

Hence, lithium phosphate is limiting reagent.

From the balanced equation,

4 mol of lithium phosphate forms 2 mol of magnesium phosphate

then,

1.511 mol of lithium phosphate forms 1.511 * 2 / 4 = 0.756 mol of magnesium phosphate.

Mass (Theoretical yield) of magnesium phosphate = 0.756 * 262.8 = 198.5 g.

Actual yield of magnesium phosphate = 125.62 g.

Therefore,

% yield of magnesium phosphate = (125.62 * 100) / 198.5 = 63.28 %

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