A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ? Number of moles of dioxygen

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n_o2 = 70.6/32 = 2.21
n_Ne = 167.5/20 = 8.375
X_o2 = 2.21/(2,21+8.375) =0 .21
X_Ne = 0.79
Partial Pressure of a Gas = Mole Fraction
Partial pressure of Oxygen = 0.21
Oartial pressure of Neon = 0.79

Partial partial pressure is equal to mole fraction x total pressure of the mixtureFind the mole fraction as given w e i g h t upon molar mass

Partial pressure = mole fraction of given gas x pure pressure.Mole fraction = Number of moles of given gas / total number of moles.Mole of dioxygen in mixture = given mass of dioxygen / molecular mass of dioxygen = 70.6 g/ 32 g = 2.2 molesMoles of Neon in mixture = given mass of neon / atomic mass of neon = 167.5 g / 20 = 8.3 moles.total number of moles = 2.2 + 8.3 = 10.5 molemole fraction of O2 = 2.2 / 10.5 = 0.2Mole fraction of neon = 8.3/ 10.5 = 0.8partial pressure of di oxygen = 0.2 × 25 = 5 barPartial pressure of neon = × 25 = 20 bar