Question

How many grams of solid barium phosphate form when 48.3 mL of 0.087 M Barium Chloride reacts with 34.0 mL of 0.155 M Sodium phosphate? What is the limiting reactant? if the percent yield is 76% how much barium phosphate is recovered?

Answer 1

3BaCl2 + 2Na3PO4 ---------> Ba3(PO4)2 + 6NaCl

moles of BaCl2 = 0.087 x 48.3/1000 = 0.0042

moles of Na3PO4 = 0.155 x 34/1000 = 0.00527

according to balanced reaction

3 moles BaCl2 reacts with 2 moles Na3PO4

0.0042 moles BaCl2 reacts with 0.0042 x 2 / 3 = 0.0028 moles Na3PO4

but we have 0.00527 moles Na3PO4 present. so Na3PO4 is exess reagent.

BaCl2 is limiting reagent.

2 moles BaCl2 forms 601.93 g Ba3(PO4)2

0.0042 moles BaCl2 forms 0.0042 x 601.93 / 2 = 12.64 g Ba3(PO4)2.

but only 76 % yield

76 = (actual yield / 12.64) x 100

actual yield / 12.64 = 0.76

actual yield = 9.61 g