Question

If 80.0 grams of oxygen gas is consumed with a stoichiometric equivalent of aluminum metal, how many grams of aluminum oxide (molar mass = 102 g/mol) would be produced?

Answer 1

Answer:

170 g

Explanation:

Moles of Oxygen gas :

Given, Mass of Oxygen = 80.0 g

Molar mass of Oxygen gas= 31.998 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{80.0\ g}{31.998\ g/mol}$

$Moles\ of\ O_2=2.5\ mol$

The reaction between Al and O₂ is shown below as:

4Al + 3O₂ ⇒ 2Al₂O₃

From the reaction,

3 moles of O₂ on reaction forms 2 moles of Al₂O₃

1 mole of O₂ on reaction forms 2/3 moles of Al₂O₃

2.5 moles of O₂ on reaction forms (2/3)*2.5 moles of Al₂O₃

Moles of Al₂O₃ = 1.6667 moles

Molar mass of Al₂O₃ = 102 g/mol

Mass = Moles * Molar mass = 1.6667 * 102 g = 170 g