Calculate the largest amount of Al2O3 that could be produced. Round your answer to the nearest 0.1 mol

Question

Solid aluminum Al

 and oxygen O2 

gas react to form solid aluminum oxide Al2O3

. Suppose you have 1.0

 mol

 of Al

 and 3.0

 mol

 of O2

 in a reactor.Calculate the largest amount of 

Al2O3

 that could be produced. Round your answer to the nearest 0.1

 mol

.

Answer 1

This question involves stoichiometry question with a need to find a limiting reactant. You will need to have a balanced equation written before being able to calculate the answer.

Balanced equation and mole ratio:

If solid Aluminum and Oxygen gas react together to form Al₂O₂, the balanced equation would look like:

4Al (s) + 3O₂ (g) --> 2Al₂O₂ (s)

with a mole ratio of 4:3:2 (4 mols of Al react with 3 mols of O to form 2 mols of 2Al2O3)

*We use this mole ratio to convert actual quantities of moles of what we have into how many moles we need to complete the reaction or how many moles we can make

Calculating Limiting Reactant:

If you have 1 mol of Al you need:

1 mol Al x 3 mol O2 = 0.75 mol O

4 mol Al

OR

If you have 3 mol O2 you need:

3 mol O₂ x 4 mol Al = 4 mol Al

3 mol O₂

Since we only have the 1 mol of Al (and not the 4 moles needed to react all 3 mol of O), Al is the limiting reactant and we'll use that to figure out how much Al₂O₃ we can make.

Calculating moles of Al₂O_3:

1 mol Al x 2 mol Al₂O₃ = 0.5 mol Al₂O₃

4 mol Al