Question

How many moles of Al2O3 can be produced from the reaction of 10.0 g of Al and 19.0 g of O2?

(a) 0.581 mol (b) 0.371 mol (c) 0.185 mol (d) 0.396 mol

Answer 1

Answer: c) 0.185mol

Step-by-step explanation:

First you need to create a balanced equation.

Al + O2 ---> Al2O3 (this is currently unbalanced)

When balanced, it is:

4Al + 3O2 ---> 2Al2O3

Then you need to find the limiting reactant (the way I do this is by taking both of them, doing dimensional analysis, and seeing which makes a smaller amount of product)

For Al:

10.0g Al * (1mol Al / 26.98g Al) * (2mol Al2O3 / 4mol Al) = 0.185mol Al2O3

For O2:

19.0g O2 * (1mol O2 / 32.00g O2) * (2mol Al2O3 / 3mol O2) = 0.396mol Al2O3

Since the amount of Al2O3 produced is smaller when using 10.0g Al, this is the limiting reactant and therefore the answer is 0.185mol Al2O3.