The limiting reactant is the chemical substance that determines the amount of product(s) that can ultimately be formed in a reaction. during the reaction, the limiting reactant is completely consumed or used up, and therefore, causes the reaction to stop. the limiting reactant can be identified through stoichiometric calculations. after comparing the results, the reactant that produces the smaller mass of product is identified as the limiting reactant. determine the excess reactant and calculate the mass of the remaining excess reactant after 20.0 grams of al and 10.0 grams of o2 react. 4 al(s) + 3 o2(g) → 2 al2o3(s) for the calculations in this module, the molar mass of an element will be rounded to the hundredths place (0.01 g).

The reactant in excess is Al (s)

The mass of remaining excess reactant is 8.76 g.

Explanation:

This is the reaction:

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

4 moles of solid aluminun reacts with 3 moles of oxygen, to make 2 moles of alumina.

Molar mass Al: 26.98 g/m

Molar mass O₂ : 32.00 g/m

Mass / Molar mass : Moles

20 g Al / 26.98 g/m = 0.741 moles

10 g O₂ / 32.00 g/m =  0.312 moles

Let's find out the limiting reactant and the reactant in excess by rules of three:

4 moles of Al reacts with 3 moles of O₂

0.741 moles of Al reacts with  (0.741 .3)/4 = 0.555 moles

Now we know that are needed 0.555 moles of O₂ to consume all the Al but we only have 0.312 moles of O₂.

The O₂ is the limiting reactant, so the Al is the reactant in excess.

Let's make another rule of three

3 moles of O₂ reacts with 4 moles of Al

0.312 of O₂ reacts with  (0.312  .4)/3 = 0.416

We need 0.416 moles to consume all the O₂ and we have 0.741 moles of Al. There's still Al to consume, that's why the Al is the reactant in excess.

When we use 0.416 moles of 0.741 moles of Al, 0.325 moles are remaining to consume all the Al.

Let's calculate the mass

0.325 moles . 26.98 g/m = 8.76 grams

the answer is c. nacl