Question

Aluminum oxide has a composition of 52.9% aluminum and 47.1% oxygen by mass. If
16.4 g of aluminum reacts with oxygen to form aluminum oxide, what mass of oxygen
reacts?

Answer 1

Answer:

The balanced chemical reaction is written as:

4Al + 3O2 = 2Al2O3

To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:

moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al

moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2

mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2

Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.

Answer 2

Answer:

The balanced chemical reaction is written as:

4Al + 3O2 = 2Al2O3

To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:

moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al

moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2

mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2

Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.