Question

13c. What is the percent yield of this reaction if 6.1 grams of P205 are produced? Write your work out on a separate piece of paper, take a picture and send via Remind or Email for the opportunity for partial credit.

Answer 1

QUESTION 13:

(a):

Balanced reaction between phosphorus and oxygen to form diphosphorus Pentoxide is given by:

4 P + 5 O₂ --------------> 2 P₂O₅

==================================================================

(b):

No.of moles of phosphorus = (15 g) / (30.97 g/mol) = 0.484 mole

No.of moles of Oxygen = (3.52 g) / (32 g/mol) = 0.11 mole

From the balanced reaction, we have:

4 moles of P reacts with 5 moles of O₂

Hence, 1 mole of P reacts with 5/4 moles of O₂

0.484 moles of P reacts with 5/4* 0.484 moles of O₂ = 0.605 mole of O₂

But we don't have that much oxygen. Hence, oxygen will be completely consumed in the reaction.

Therefore, Oxygen is the limiting reagent.

So calculations must be based on moles of O₂.

From the balanced reaction, we have:

5 moles of O₂ reacts to produce 2 moles of P₂O₅

Hence, 1 mole of O₂ reacts to produce 2/5 moles of P₂O₅

Therefore, 0.11 mole of O₂ reacts to produce 2/5*0.11 mole of P₂O₅ = 0.044 mole of  P₂O₅

Therefore, Mass of product ( P₂O₅) formed = (0.044 mole) x (141.94 g/mol) = 6.245 g P₂O₅ ----------(ANSWER)

==================================================================

(c):

Percent yield = (actual yield) / (theoretical yield) * 100%

Percent yield = (6.1 g) / (6.245 g) * 100%

Percent yield = 0.9767 * 100%

Percent yield = 97.67 % ----------(ANSWER)