Question

What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus

Answer 1

`n_(P_2O_5)^(max)=1.4molP_2O_5`

Step-by-step explanation:

Hello!

In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

`2P+(5)/(2)O_2\rightarrow P_2O_5`

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

`n_(P_2O_5)^(max)=112 gO_2*(1molO_2)/(32.00gO_2)*(1molP_2O_5)/(5/2molO_2)\\\\ n_(P_2O_5)^(max)=1.4molP_2O_5`

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