Using the reaction, #4Al + 3O_2 -> 2Al_2O_3#, how do you identify the limiting reactant in each of the following: 0.25 mol Al and .40mol #O_2# and 58.5g Al and 98.0g #O_2#?

1 Answer
Andrea B.
Jul 19, 2017

Al

Explanation:

The limiting reactant is the reactant that is present in less amount. 4 mol of Al need to react completely of 3 mol of Oxygen (according with the reaction you have written). Making a proportion, you have 0,25 mol of Al so you need 0,25 x 3/4 = 0,1875 mol di Oxygen. You have 0,4 of that, therefore the limiting reactant is Al. in the second case you have #(58,5g)/(27g/(mol)) =2,17 mol Al# and #(98g)/ (32g/(mol))= 3,06 mol O_2# . Making a proportion, for 2,17 mol of Al you need 1,62 mol of Oxygen < 3,06. The limiting reactant is Al again

Related questions
See all questions in Limiting Reagent
Impact of this question
3500 views around the world
You can reuse this answer
Creative Commons License