$A$ neon-dioxygen mixture contains $70.6 \ g$ dioxygen and $167.5 \ g$ neon. If the total pressure of the mixture of gases in the cylinder is $25 \ bar$,what are the partial pressures of dioxygen and neon in the mixture?

(N/A) Number of moles of dioxygen $(n_{O_2})$ $= \frac{70.6 \ g}{32 \ g \ mol^{-1}} = 2.21 \ mol$.
Number of moles of neon $(n_{Ne})$ $= \frac{167.5 \ g}{20 \ g \ mol^{-1}} = 8.375 \ mol$.
Total moles $(n_{total})$ $= 2.21 + 8.375 = 10.585 \ mol$.
Mole fraction of dioxygen $(x_{O_2})$ $= \frac{2.21}{10.585} = 0.21$.
Mole fraction of neon $(x_{Ne})$ $= \frac{8.375}{10.585} = 0.79$.
Partial pressure of a gas $= \text{mole fraction} \times \text{total pressure}$.
Partial pressure of dioxygen $= 0.21 \times 25 \ bar = 5.25 \ bar$.
Partial pressure of neon $= 0.79 \times 25 \ bar = 19.75 \ bar$.