Question

When 3.54 grams of phosphorus react with excess oxygen gas to produce diphosphorus pentoxide, the percent yield is 84.1%. What was the actual yield of the reaction? 4P 5O2 yields 2P2O5

Answer 1

Answer:

Actual yield of P2O5 = 13.6 g

Explanation:

The given reaction is:

$4P + 5O2 \rightarrow 2P2O5$

It is mentioned that O2 gas is in excess hence phosphorous (P) is the limiting reagent which will dictate the yield of the product.

Atomic wt of P = 30.97 g/mol

$Moles\ of\ P\ present = \frac{Mass}{At. mass} =\frac{3.54g}{30.97g/mol} =0.114 moles$

Based on the reaction stoichiometry:

4 moles of P produces 2 moles of P2O5

Therefore:

$moles\ P2O5\ produced = \frac{1}{2} * moles\ of\ P$

$=\frac{1}{2} *0.114=0.057moles$

Molecular mass of P2O5 =141.94 g/mol

$Mass\ of\ P2O5\ produced = moles*mol.wt = 0.114moles*141.94g/mol=16.18g$

Therefore theoretically if the yield is 100% then the amount of P2O5 produced =16.18 g

Since the yield is only 84.1%, the P2O5 produced would be:

= 0.841*16.18 = 13.6 g

Answer 2

The atomic mass for phosphorus is 31. So the molar mass of P is 3.54/31=0.114 mol. And according to the reaction equation, the ratio of coefficient is equal to the mole number. So the theoretical mass of P2O5 is 8.094 g. The actual yield is 6.807 g.