If you are given 8.8x10^23 atoms of aluminum, how many liters of fluorine gas will you need to react with

Answer 1

2Al(s) + 3F₂(g) ==> 2AlF₃(s)

moles of Al used = 8.8x10²³ atoms x 1 mole/6.02x10²³ atoms = 1.46 moles Al

moles of F₂(g) required = 1.46 moles Al x 3 moles F₂/2 mole Al = 2.19 moles F₂ needed

The volume (liters) of F₂ required will depend on the temperature and pressure of the system. If we assume STP (atm and 273K), and assume F₂ behaves as an idea gas, then ...

2.19 moles F₂(g) x 22.4 L/mole = 49 liters F₂ gas required (to 2 signifiant figures)