Question

An unknown amount of Al203 decomposed producing 215 g of solid aluminum. 2Al2O3=4Al+3O2 How many grams of oxygen gas should be produced

Answer 1

Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• Al₂O₃: 2 moles
• Al: 4 moles
• O₂: 3 moles

Being the molar mass of each compound:

• Al₂O₃: 102 g/mole
• Al: 27 g/mole
• O₂: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• Al₂O₃: 2 moles* 102 g/mole= 204 grams
• Al: 4 moles* 27 g/mole= 108 grams
• O₂: 3 moles* 32 g/mole= 96 grams

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

$mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}$

mass of oxygen= 191.11 grams

191.11 grams of oxygen gas should be produced.