Question

between 0.500g of aluminum and excess oxygen in the atmosphere,which is a limiting reactant

Answer 1

To determine the limiting reactant between 0.500g of aluminum and excess oxygen in the atmosphere, we need to compare the amounts of each reactant and determine which one will be completely consumed in the reaction.

First, we need to convert the mass of aluminum to moles. The molar mass of aluminum is 26.98 g/mol.

0.500g Aluminum * (1 mol/26.98g) = 0.01852 mol Aluminum

Next, we need to consider the balanced chemical equation for the reaction between aluminum and oxygen:

4Al + 3O2 → 2Al2O3

From the balanced equation, we can see that it requires 4 moles of aluminum for every 3 moles of oxygen to form 2 moles of aluminum oxide.

So, we can calculate the amount of oxygen required to react with 0.01852 moles of aluminum:

0.01852 mol Aluminum * (3 mol Oxygen / 4 mol Aluminum) = 0.01389 mol Oxygen

Since there is excess oxygen in the atmosphere, which means there is more oxygen available than needed, the limiting reactant is the aluminum.

Therefore, in this reaction, 0.500g of aluminum is the limiting reactant.