Question

One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 19.9 g of Ca₃(PO₄)₂ reacts with 54.3 g of H₂SO₄, what is the percent yield if 10.9 g of H₃PO₄ is formed via the UNBALANCED equation below?
Ca₃(PO₄)₂ (s) + H₂SO₄ (aq) → H₃PO₄ (aq) + CaSO₄ (aq)

Answer 1

57.5 % is the percent yield if 10.9 g of H₃PO₄ is formed.

Explanation:

Balanced equation for the reaction:

Ca₃(PO₄)₂ (s) + 2H₂SO₄ (aq) → H₃PO₄ (aq) + 2CaSO₄ (aq)

data given:

mass of Ca₃(PO₄)₂ = 19.9 grams

mass of  H₂SO₄, = 54.3 grams

mass of H₃PO₄ produced = 10.9 grams (actual yield)

percent yield=?

atomic mass of Ca₃(PO₄)₂  = 310.17 grams/mole

atomic mass of H₂SO₄ = 98.07 grams/mole

number of moles is calculated as:

number of moles  = $\frac{mass}{atomic mass of one mole}$

putting the values in the above equation:

for Ca₃(PO₄)₂  = $\frac{19.9}{310.17}$

                        = 0.064 moles

moles of H₂SO₄ = $\frac{54.3}{98.07}$

                            = 0.55 moles

from the reaction, it can be found that the limiting reagent is Ca₃(PO₄)₂

1 mole of Ca₃(PO₄)₂ reacts to form 1 mole of phosphoric acid

0.064 moles of  Ca₃(PO₄)₂ will produce x moles of phosphoric acid

0.064 moles of phosphoric acid produced

mass = number of moles x atomic mass of  H3PO4

             =0.064 x 97.994

           = 6.27 grams (theoretical yield)

FORMULA FOR PERCENT YIELD:

percent yield = $\frac{actual yield}{theoretical yield}$ x 100

                      = $\frac{6.27}{10.9}$ x 100

                       = 57.5 %