Question

1. Phosphorous reacts with bromine to form phosphorous tribromide. If 35.0 grams of bromine

Answer 1

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 %