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

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