Question

If an enclosure of 0.462 L has a partial pressure of O2 of 3.6×10?6 torr at 28 ?C, what mass of magnesium will react according to the following equation? 2Mg(s)+O2(g)?2MgO(s)

Answer 1

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 3.6×10?6 torr , change this to atm:

3.6×10?6 torr / 760 = 4.74 x 10^-9 atm

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

moles = 4.74 x 10^-9 * 0.462 / (0.082 * 301.15) = 8.86 x 10^-11 moles of oxygen

1 mole of O2 requires 2 moles of magnesium

moles of magnesium required = 2 * 8.86 x 10^-11 = 1.77x10^-10 moles

atomic mass of magnesium = 24.3

mass of magnesium = 1.77x10^-10 moles * 24.3 = 4.31 x 10^-9 grams of magnesium

*hope it helps =)