4Al(s)+3O2(g)—>2Al2O3(s) the reaction is started by mixing 10.0g of Al and 10.0g of O2. What is the limiting reaction and how much Al2O3 is produced?

A. Al is the limiting reactant.

B. 18.89g of Al2O3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

4Al(s) + 3O2(g) —>2Al2O3(s)

Next, let us determine the masses of Al and O2 that reacted and the mass of Al2O3 produced from the balanced equation. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al from the balanced equation = 4 x 27 = 108g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 = 102g/mol

Mass of Al2O3 from the balanced equation = 2 x 102 = 204g

Summary:

From the balanced equation above, 108g of Al reacted with 96g of O2 to produce 204g of Al2O3.

A. Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above, 108g of Al reacted with 96g of O2.

Therefore, 10g of Al will react with = (10 x 96)/108 = 8.89g of O2.

From the calculations made above, we can see that lesser mass of O2 is needed to react completely with 10g of Al given. Therefore, Al is the limiting reactant.

B. Determination of the mass of Al2O3 produced from the reaction.

In this case, the limiting reactant will be used because it will produce the maximum yield of the reaction as all of it is used up in the reaction. The limiting reactant is Al. The mass of Al2O3 produced by the reaction can be obtain as follow:

From the balanced equation above, 108g of Al reacted to produce 204g of Al2O3.

Therefore, 10g of Al will react to produce = (10 x 204)/108 = 18.89g of Al2O3.

Therefore, 18.89g of Al2O3 is produced from the reaction.