# 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 $\frac{58 , 5 g}{27 \frac{g}{m o l}} = 2 , 17 m o l A l$ and $\frac{98 g}{32 \frac{g}{m o l}} = 3 , 06 m o l {O}_{2}$ . Making a proportion, for 2,17 mol of Al you need 1,62 mol of Oxygen < 3,06. The limiting reactant is Al again