Question

2 Aluminum oxidizes according to the following equation: 4A1 + 302 → 2Al2O,

[a] Powdered Al (0.048 myl) is placed into a container containing 0.030 mol O.. What is the limiting reactant?

[b] How many moles of the excess reoctant remaine?​

Answer 1

The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed. In this case, powdered Al is the excess reactant and O2 is the limiting reactant. Since O2 is the limiting reactant, all of it will be consumed.

Step-by-step explanation:

The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the amount of product that can be formed. In order to determine the limiting reactant, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation.

a) We have 0.048 mol of powdered Al and 0.030 mol of O2. From the balanced equation, we can see that 4 moles of Al react with 3 moles of O2. Therefore, we need to calculate how many moles of Al would react with 0.030 mol of O2. This can be done using stoichiometry:

Moles of Al = (0.030 mol O2) × (4 mol Al / 3 mol O2) = 0.040 mol Al

Since we have 0.048 mol of Al, which is greater than 0.040 mol, Al is the excess reactant and O2 is the limiting reactant.

b) To find out how many moles of the excess reactant remain, we need to calculate the moles of O2 that reacted. Since we know that 0.030 mol of O2 reacted, and O2 is the limiting reactant, all of the O2 will be consumed. Therefore, there will be no moles of O2 remaining.