# What is the maximum amount in moles of P_2O_5 that can theoretically be made from 186 g of P_4 and excess oxygen?

Jun 28, 2018

Well, we need some stoichiometry...

#### Explanation:

$\frac{1}{2} {P}_{4} \left(s\right) + \frac{5}{2} {O}_{2} \left(g\right) \rightarrow {P}_{2} {O}_{5} \left(s\right)$

$\text{Moles of phosphorus}$ $=$ $\frac{186 \cdot g}{31.00 \cdot g \cdot m o {l}^{-} 1} = 6 \cdot m o l \cdot \text{phosphorus atoms} \ldots$...

And so we can make AT MOST $3 \cdot m o l$ ${P}_{2} {O}_{5}$...

We need $15 \cdot m o l \cdot \text{oxygen ATOMS}$ for equivalence...

$15 \cdot m o l \times 16.00 \cdot g \cdot m o {l}^{-} 1 = 240 \cdot g$...

And AT MOST we can make $\text{THREE MOLES}$ of ${P}_{2} {O}_{5}$...a mass of $3 \cdot m o l \times 112 \cdot g \cdot m o {l}^{-} 1 = 336 \cdot m o l$..

Of course, ${P}_{2} {O}_{5}$, is actually ${P}_{4} {O}_{10}$..but we can use this formulation for determination of equivalence...