Answer:
70.6 %
Explanation:
First step, we define the reaction:
2P + 3Br₂ → 2PBr₃
We determine the moles of reactant:
35 g . 1mol / 159.8 g = 0.219 moles
We assume, the P is in excess, so the bromine is the limiting reagent.
3 moles of Br₂ can produce 2 moles of phophorous tribromide
Then, 0.219 moles may produce (0.219 . 2) /3 = 0.146 moles of PBr₃
We convert moles to mass:
0.146 mol . 270.67 g /mol = 39.5 g
That's the 100 % yield reaction, also called theoretical yield. The way to determine the % yield is:
(Yield produced / Thoeretical yield) . 100
(27.9 / 39.5) . 100 = 70.6 %
