Question

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`?

Answer 1

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