Answer:

To answer these questions, we need to first determine the limiting reactant.

Using the balanced chemical equation, we can find the mole ratio of Na to O2 as:

4 mol of Na : 1 mol of O2

Then, we can calculate the amount of Na and O2 in moles:

moles of Na = 5.00 g / 23.00 g/mol = 0.217 mol

moles of O2 = 5.00 g / 32.00 g/mol = 0.156 mol

The limiting reactant is the one that is completely consumed in the reaction. To determine which one that is, we compare the mole ratio of the reactants to see which one is smaller. In this case, we see that there is a smaller amount of O2, so it is the limiting reactant.

a. The stoichiometry of the balanced chemical equation tells us that every 1 mol of O2 that reacts can produce 2 mol of Na2O. Therefore, the maximum amount of product that can be formed is:

moles of Na2O = 2 * moles of O2 = 2 * 0.156 mol = 0.312 mol

The mass of Na2O that can be formed is:

mass of Na2O = moles of Na2O * molar mass of Na2O

mass of Na2O = 0.312 mol * 61.98 g/mol = 19.27 g

So, the maximum amount of product that can form is approximately 19.27 g.

b. As we calculated earlier, the limiting reactant is O2.

c. To determine the amount of excess reactant, we can use the mole ratio of Na to O2 from the balanced chemical equation. Since every 4 moles of Na react with 1 mole of O2, we can calculate the amount of Na that would be required to react completely with 0.156 mol of O2:

moles of Na required = 4 * 0.156 mol = 0.624 mol

The amount of excess Na is then:

moles of Na in excess = moles of Na present - moles of Na required

moles of Na in excess = 0.217 mol - 0.624 mol = -0.407 mol

The negative value indicates that there is no excess Na left over, and that all of the Na reacted with O2.