Question

How many moles of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$, will contain $0.25$ mole of Oxygen atoms?

Answer 1

Moles of Oxygen in ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$:

• One mole of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$ contains: three moles of $\mathrm{Mg}$ atom, two moles of $\mathrm{P}$ atom, and eight moles of $\mathrm{O}$atoms.
• Eight moles of Oxygen atoms are present in one mole of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$.
• So, $0.25$ moles of Oxygen atom is present in $\mathrm{x}$ moles of${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$.

$$\begin{gathered} 8molesofOxygen\to 1moleofM{g}_3{\left({\mathrm{PO}}_4\right)}_2 \\ 0.25molesofOxygen\to xmolesofM{g}_3{\left({\mathrm{PO}}_4\right)}_2 \\ x=\frac{0.25\times 1}{8}=0.03125moles \end{gathered}$$

Hence, $0.01325$ mol of ${\mathrm{Mg}}_3{\left({\mathrm{PO}}_4\right)}_2$ contains $0.25$moles of Oxygen.