percent yield weird problem?

Question

Percent yield for the reaction PCl3(l) + Cl2(g) -> PCl5(s) is 83.2%. What mass of PCl5 would be expected from the reaction of 56.7g of PCl3 with excess chlorine?

Answer 1

Okay, so you're given the stoichiometry of the reaction, and told that the yield is going to be 83.2%.
Then it says you're starting with 56.7g of PCl3 (one of the starting materials), and asked how much product PCl5 you should expect, given that the reaction will proceed in 83.2% yield.
From a periodic table, we know that the formula weight of PCl3 is 137.3 g/mole. We're starting with 56.7 g, which is 56.7g / 137.3 g/mol = 0.41 moles of PCl3. If the reaction went in 100% yield, then we should get 0.41 moles of PCl5 product, right? Since the stoichiometry is 1:1 in the reaction, one molecule of PCl3 plus one molecule of Cl2 equals one molecule of PCl5.
But we know the reaction is only proceeding in 83.2% yield, so we should therefore expect 0.41 moles x 0.832 = 0.34 moles of PCl5. From a periodic table we can figure out that PCl5 weighs 208.2 g/mole. Therefore 0.34 moles x 208.2 g/mole = 71.5 g of PCl5.
The answer is 71.5 g.
Good luck!

Answer 2

Atomic weights: P=31 Cl=35.5 PCl3=137.5 PCl5=208.5

56.7gPCl3 x 1molPCl3/137.5gPCl3 x 1molPCl5/1molPCl3 x 208.5gPCl5/1molPCl5 x 83.2% = 71.5 g PCl5

Answer 3

The theoretical yield is the ensuing conversion of the reaction assuming suitable circumstances. the honestly yield is what you purchased once you bodily do the test. The p.c. yield is the honestly yield over theoretical yield (how close to to *suitable* your test became) multipled via one hundred to offer a proportion. Such that if the reaction became meant to yield 3g of product (theoretical yield) yet you in straightforward terms get a million.5g of product (honestly yield) you've a p.c. yield of a million.5g/3g = 0.5 * one hundred% = 50%.