Question

Consider the following balanced equation Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) -------> 3CaSO₄ (s) + 2H₃PO₄ (aq)

What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H₂SO₄ by mass)?

Answer 1

Ans. Ca₃(PO₄)_2(s) + 3 H₂SO_4(aq) -----------> 3 CaSO_4(s) + 2 H₃PO₄

Number of moles of calcium phosphate in 1.0 kg sample

                        = mass of Ca. phosphate in g / molecular mass

                        = 1000 g / 310.17 g mol^-1

                        = 3.2240 moles

Number of moles of calcium phosphate in 1.0 kg sample

            Given, H₂SO₄ is 98% by mass. That is the actual weight of H₂SO₄ in sample = 98 % of 1 kg = 0.98 kg = 980 gram

                        = mass of H₂SO₄ in g / molecular mass

                        = 980 g / 98.072 g mol^-1

                        = 9.992 moles

According to stoichiometry of balance reaction, 1 mol Ca.P requires 3 moles H₂SO₄.

Thus, total moles of H₂SO₄ required by 3.2240 moles Ca.P =

                        = 3 x 3.2240 moles = 9.672 moles

Thus, H₂SO_4­ is present in excess and all Ca.P will be consumed up.

Moreover, all the products are produced from 3.2240 moles of Ca.P

A. moles of CaSO4 = 3 x moles of Ca.P

                                    = 3 x 3.2240 moles = 9.672 moles

Mass of CaSO4 = moles x molecular mass

                        = 9.672 moles x 136.13 g mol^-1 = 1316.69 g

B. moles of H3PO4 = 2 x moles of Ca.P

                                    = 2 x 3.2240 moles = 6.448 moles

Mass of H3PO4 = moles x molecular mass

                        = 6.448 moles x 97.994 g mol^-1 = 631.86 g