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#?
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