Question

One of the major uses of sulfuric acid is the production of phosphoric acid and calcium sulfate. The reaction is: Ca3(PO4)2(s) + 3H2SO4(aq) → 3CaSO4(s) + 2H3PO4(aq) What mass of concentrated sulfuric acid (98.0 % H2SO4 by mass) must be used to react completely with 155.9 g of Ca3(PO4)2?

Answer 1

Ca₃(PO4)₂(s) + 3H₂SO₄(aq) → 3CaSO₄(s) + 2H₃PO₄(aq)

Mass of  Ca₃(PO4)₂ =155.9 g

Moles of Ca₃(PO4)₂=155.9 g / 310 g mol^-1 [ Moar mass of Ca₃(PO4)₂=310 g mol^-1 ]

=0.5029 moles

From the balanced chemical equation

1 mole of  Ca₃(PO4)₂ requires 3 moles of H₂SO₄

Therefore,0.5029 moles of  Ca₃(PO4)₂ will require = 3 x 0.5029 moles of H₂SO₄

= 1.5087 moles of H₂SO₄

Molar mass of H₂SO₄ = 98 g/mol

Mass of 1.5087 moles of H₂SO₄ = 1.5087 moles x 98 g/mol

=147.8526 g

This is the mass of 100% sulfuric acid required.

Mass of 98% sulfuric acid required = 147.8526 g/ 0.98 = 150.87 g