# Chem Problem: If an enclosure of 0.442 L has a partial pressure of O2 of 3.5×10−6 torr at 28 ∘C, what mass of magnesium will react?

If an enclosure of 0.442 L has a partial pressure of O2 of 3.5×10−6 torr at 28 ∘C, what mass of magnesium will react according to the following equation?

2Mg(s)+O2(g)→2MgO(s)

Relevance

The first thing is calculate the number of moles using the ideal gas equation

PV= moles * R*T

P is pressure

V is volume

R is gas constant

T is temperature

moles = Pressure * volume / (gas constant * Temperature)

temperature is 28 C, change this to kelvin

Temperature kelvin = 28 + 273.15 = 301.15 K

Pressure is 3.5×10?6 torr , change this to atm:

3.5×10?6 torr / 760 = 4.60 x 10-9 atm

Gas constant "R" = 0.082 atm L / kelvin mole

moles = 4.60 x 10-9 * 0.462 / (0.082 * 301.15) = 8.57 x 10-11 moles of oxygen

1 mole of O2 requires 2 moles of magnesium

moles of magnesium required = 2 * 8.57 x 10-11 = 1.71x10-10 moles

atomic mass of magnesium = 24.3

mass of magnesium = 1.71x10-10 moles * 24.3 = 4.16 x 10-9 grams of magnesium

*hope it helps =)