Question

30 POINTS! ----- How many moles of oxygen gas are needed to completely react with 145 grams of aluminum? Report your answer with 3 significant figures. ____ mole O2.

Answer 1

First a balanced reaction equation must be established:

`4Al _((s)) + 3 O_(2) _((g))` →
`2 Al_(2) O_(3)`

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols

Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]

∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)

Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.