Question

How many grams of aluminum oxide are produced according to the reaction below given that you start with 13.2 grams of Al and 13.2 grams of O2?

Reaction: 4AI + 3 02 → 2A1203

Answer 1

Answer: 24.87g Al2O3

Explanation: Aluminum is our limiting reagent

`13.2g Al * (1 mol Al)/(26.98 g) * (2 mol Al2O3)/(4 mol Al) = 0.244 mol Al2O3`

First of all we need to convert the grams of aluminium to moles, then use the molar fraction of the balanced equation (4 moles of aluminium equals 2 of aluminum oxide).

If we made the same procedure with the oxigen we get a 0.273 mol of Al2O3, therefore the O2 is the excess reagent.

The last step is convert the moles of the limiting reagent to grams

`0.244 mol Al_(2) O_(3) * (101.96g)/(1mol Al2O3) = 24.87g Al2O3`

And that´s it!