Question

In Part A, you found the amount of product (2.60mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the amount of product (2.20mol P2O5 ) formed from the given amount of oxygen and excess phosphorus.

Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus andoxygen.

Express your answer to three significant figures and include the appropriate units.

Answer 1

The reaction is

2P(s) + 2.5O2(g) ---> P2O5

so as per balanced equation, two mole of P reacts with 2.5 moles of O2 to give one mole of P2O5.

As mentioned, in part A the amount of P2O5 formed is 2.6 moles when we have taken excess O2, so here the amount of Phosphorous must be 2X2.6 moles = 5.2 moles

In part B, the amount of P2O5 formed is 2.2 moles when we have taken P in excess, so the moles of O2 taken = 2.5 X 2.2 moles

Moles of O2 used = 5.5 moles

Part (C)

The moles of O2 used = 5.5 moles

The moles of P used =5.2 moles

So here O2 is the limiting reagent, as for 5.2 moles of P we need = 6.5 moles of O2

As per balanced equation,

2.5 moles of O2 gives one mole of P2O5

one mole of O2 will give 1/2.5 moles of P2O5

So 5.5 moles of O2 will give 5.5 / 2.5 moles of P2O5 = 2.2 moles of P2O5

Molecular weight of P2O5 = 142g / mole

amount of P2O5 obtained = Moles X molecular weight = 2.2 X 142 = 312.4 grams