1. Phosphorous reacts with bromine to form phosphorous tribromide. If 35.0 grams of bromine reacted and 27.9 grams of phosphorous tribromide are actually formed.
what is the percent yield of phosphorous tribromide (PBr3)?
2 P + 3 Br2 ---> 2 PBr3
(Note that the molar masses: P=30.97g, Br = 79.90g) Show your steps
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Please help me

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 %

to make a supersaturated solution, make a saturated solution of sugar by adding 360 grams of sugar to 100 ml of water at 80 degrees celsius. when the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water should only dissolve 210 grams of sugar.