Question

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

Answer 1

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.