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Sati [7]
3 years ago
8

In Part A, you found the number of moles of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxyge

n. In Part B, you found the number of moles of product (1.40 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus. Now, determine the number of moles of P2O5 is produced from the given amounts of phosphorus and oxygen
Chemistry
1 answer:
n200080 [17]3 years ago
5 0

Answer:

1.40 moles.

Explanation:

The balanced chemical equation for the reaction of Phosphorus and Oxygen is as follows -

4P + 5O_2 = 2P_2O_5

In Part A, the oxygen was taken in excess. So Phosphorus will be the limiting reagent.

Since, 2 moles of P_2O_5 is formed by 4 moles of P

So, for 1.8 moles of  P_2O_5 amount of required moles of P = \frac{4}{2} \times 1.80 = 3.60

In Part B, the phosphorus was taken in excess so oxygen will be the limiting reagent.

Since, 2 moles of P_2O_5 is formed by 5 moles of oxygen

So, for 1.40 moles of P_2O_5 moles of O_2 required = \frac{5}{2} \times 1.40 = 3.50

Thus as of now we have 3.60 moles of P_2O_5 and 3.50 moles of O_2.

As in the reaction of formation of P_2O_5, oxygen is the limiting reagent.

So the moles of P_2O_5 formed by the 3.50 moles of oxygen will be

P_2O_5 = \frac{2}{5} \times 3.50 = 1.40 moles.

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