Answer:
The percent yield reaction is 64.3%
Explanation:
This is the ballanced reaction
Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)
Let's determine the moles of our reactants:
Mass / Molar mass = Mol
26.8 g / 310.18 g/m = 0.0864 moles of phosphate.
54.3 g / 98.06 g/m = 0.554 moles of sulfuric
1 mol of phosphate reacts with 3 mol of sulfuric so
0.0864 mol of PO₄⁻³ will react with (0.0864 .3)/1 = 0.259 moles
I have 0.554 of sulfuric, so this is the reactant in excess.
The limiting reagent is the Phosphate.
1 mol of PO₄⁻³ produces 2 mol of phosphoric
0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles
Mol . molar mass = Mass
0.173 m . 97.98g/m = 16.95 g (This is the theoretical yield)
Percent yield = (Produced / Theoretical) .100
(10.9 g / 16.95 g) . 100 = 64.3 %
