For the equation P4 (s) + 5 O2 (g) → P4 O10 (s), if 3 mol of phosphorous react with 10 mol of oxygen, the theoretical yield of phosphorous (V) oxide will be

Theoretical yield of P4O10 is 568g

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

P4 (s) + 5O2 (g) → P4O10 (s)

Step 2:

Determination of the limiting reactant.

From the balanced equation above, 1 mole of P reacted with 5 moles of O2.

Therefore, 3 moles of P will react with = 3 x 5 = 15 moles of O2.

We can see that a higher amount of O2 than what was given is needed to react with 3 moles of P. Therefore, O2 is the limiting reactant.

Step 3:

Determination of the theoretical yield of P4O10.

In this case the limiting reactant is used as it will produce the maximum yield of the reaction. The limiting reactant is O2. The theoretical yield of P4O10 is obtained as follow:

From the balanced equation above, 5 moles of O2 produced 1 mole P4O10.

Therefore, 10 moles of O2 will produce = (10 x 1) / 5 = 2 moles of P4O10.

Next, we'll convert 2 moles of P4O10 to grams to obtain the desired result. This is illustrated below:

Number of mole of P4O10 = 2 moles

Molar Mass of P4O10 = (31x4) + (16x10 = 124 + 160 = 284g

Mass of P4O10 = ?

Mass = mole x molar Mass

Mass of P4O10 = 2 x 284

Mass of P4O10 = 568g

Therefore, the theoretical yield of P4O10 is 568g.