Consider a situation in which 112 g of P4 are exposed to 112 g of O2 . 1- What i
ID: 724322 • Letter: C
Question
Consider a situation in which 112 g of P4 are exposed to 112 g of O2 .1- What is the maximum amount of P2O5 that can theoretically be made from 112 g of O4 and excess oxygen?
2- What is the maximum amount of P2O5 that can theoretically be made from 112 g of O2 and excess phosphorus?
Explanation / Answer
sol. write a balanced equation P4 + 5O2 --------> 2P2O5 work out moles of each reagent Then work out moles of product possible if all of the reagent were used up. The one that can produce the least product is the limiting reagent. moles P4 = mass / molar mass = 112 g / 123.88 g/mol = 0.9041 mol P4 1 mole P4 reacts to give 2 moles P2O5 So 0.9041 moles will give (2 x 0.9041) mol P2O5 = 1.81 mol P2O5 possible from 112 g P4 moles O2 = 112 g / 32.00 g/mol = 3.50 mol 5 mole O2 reacts to give 2 mole P2O5 So 1 mol O2 can produce 2/5 mole P2O5 thus 3.50 mol O2 can form up to (2/5 x 3.50 ) mole P2O5 = 1.40 mole P2O5 possible from 112 g O2 The O2 can only form 1.40 moles P2O5, So O2 is the limiting reagent (because the amount of P4 provided is enough to produce 1.81 mol P2O5, but it can't because there is not enough O2) The theoretical yield is the maximum possible yield of product given the quantities of reagent provided. So it is amount od P2O5 produced if all the limiting reagent is used up. From above this is 1.40 moles P2O5 2. The balanced equation shows that 4 moles Cr meeds 3 moles O2 to fully react So 1 mole Cr will need 3/4 moles O2 to fully react Thus the O2 is in excess, because you only need 0.75 moles O2 to react with the 1 mol Cr but you are given 1 mol O2, so there will be 0.25 mol