In: Chemistry
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)?
Ans. Ca3(PO4)2(s) + 3 H2SO4(aq) -----------> 3 CaSO4(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, H2SO4 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