Question

13. What is the experimental yield (in grams) of the solid product when the percent yield is 93.4 % when 9.946 g of barium chloride reacts in solution with excess sodium phosphate?

BaCl₂(aq) + Na₃PO₄(aq) --> Ba₃(PO₄)₂(s) + NaCl(aq) [unbalanced]

Answer 1

molar mass of barium chloride = 208.23 gm /mol

9.946 gm = 9.946 / 208.23 = 0.0477 mole

standard reaction predicts 3 mole of barium chloride produces 1 mole of barium phosphate as solid product

so, 0.0477 mole produces = 0.0159 mole

molar mass of barium phosphate = 601.93 gm/mole

hence, theoretical yield = 0.0159 mole*601.93 gm/mole = 9.57 gm

% yield = (experimental yield / theoretical yield )*100

or, 93.4 = (experimental yield / 9.57 )*100

or, experimental yiel =8.94 gm (answer)