Question

Consider a situation in which 211 g of P4 are exposed to 240 g of O2. The balanced chemical equations is P4+5O2=2P2O5. A) What is the maximum amount in moles of P2O5 that can theoretically be made from 211 g of P4 and excess oxygen? B)What is the maximum amount in moles of P2O5 that can theoretically be made from 240 g of O2 and excess Phosphorus? C)In part A, you found the amount of product (3.40 mol of P2O5) formed from the given amount of phosphorus and excess oxygen. In part B, you found the amount of product (3.00 mol P2O5) formed from the given amount of oxygen and excess phosphorus. D)What is the percent yield if the actual yield from this reaction is 260 g?

Answer 1

A)

P4 + 5O2 --------> 2P2O5

here P4 is the limiting reagent.
The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

moles P4 = mass / molar mass
= 211 g / 123.88 g/mol
= 1.70 mol P4

1 mole P4 reacts to give 2 moles P2O5
So 0.9041 moles will mol P2O5

=1.70 mol P4 * 2 mole P2O5/ 1 mole P4

= 3.40 mole

B)

P4 + 5O2 --------> 2P2O5

here O2 is the limiting reagent.
The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

moles O2 = mass / molar mass
= 240 g / 32.00 g/mol
= 7.500 mol O2

5 mole O2 reacts to give 2 mole P2O5
So 1 mol O2 can produce 2/5 mole P2O5
thus 7.50 mol O2 can form up to (2/5 x 7.50 ) mole P2O5
= 3.00 mole P2O5 possible from 240 g O2

C)In part A, you found the amount of product (3.40 mol of P2O5) formed from the given amount of phosphorus and excess oxygen. In part B, you found the amount of product (3.00 mol P2O5) formed from the given amount of oxygen and excess phosphorus.