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

Answer:

$n_{P_2O_5}^{max}=1.4molP_2O_5$

Explanation:

Hello!

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

$2P+\frac{5}{2}O_2\rightarrow P_2O_5$

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:

$n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\ n_{P_2O_5}^{max}=1.4molP_2O_5$

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