Chapter 9, Problem 6A

Interpretation Introduction

Interpretation:

The mole ratio used to calculate the number of moles of oxygen gas required to react completely with a given number of moles of aluminum metal needs to be determined.

Also, the mole ratio used to calculate the number of moles of the product expected if a given number of moles of aluminum metal reacts completely needs to be determined.

Concept Introduction:

The mole ratio should use to calculate how many moles of oxygen gas would be needed to react completely with a given number of moles of aluminum metal is (3 mol O₂ / 4 mol Al).

Answer to Problem 6A

The mole ratio of Al : O₂ = 4 : 3

The mole of the limiting reactant to the product = 4:2 or 2:1

Use the mole coefficient of Al since it is completely consumed in the reaction and is thus the limiting reactant.

Therefore, use the mole ratio of Al:Al₂O₃

Explanation of Solution

The mole ratio should use to calculate how many moles of oxygen gas would be needed to react completely with a given number of moles of aluminum metal is (3 mol O₂ / 4 mol Al).

Here, 4 mol Al is used because it is stated in the balanced chemical equation that 4 moles Al react with 3 moles of O₂. So, put 4 moles Al in the denominator so that the units will cancel with the given value and mol O₂ will be left.

  $\text{mol Al×}\frac{{\text{3 mol O}}_{\text{2}}}{\text{4 mol Al}}{\text{=mol O}}_{\text{2}}$

The mole ratio uses to calculate the number of moles of product that would be expected if a given number of moles of aluminum metal reacts completely is (2 mol Al₂O₃ / 4 mol Al).

Here, used 4 mol Al because it is stated in the balanced chemical equation that 4 moles Al is needed to produce 2 mol Al₂O₃.

So, put the 4 mol Al in the denominator so that the units will cancel with the given and mol Al₂O₃ will be left as shown below:

  $\text{mol Al×}\frac{{\text{2 mol Al}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{4 mol Al}}{\text{=mol Al}}_{\text{2}}{\text{O}}_{\text{3}}$