Question

Given 25.0 g of O2, how many grams of the product Al2O3 could be produced?

Answer 1

The balanced chemical reaction between aluminum and oxygen is as follows:

4Al(s) + 3O₂(g) $\rightarrow$ 2Al₂O₃(s)

Mass of oxygen(O₂) = 25.0 g

Determine the number of moles of oxygen using the given mass and molar mass of oxygen as follows:

The molar mass of O₂ = 31.9988 g/mol

= 25.0 g O₂ x ( 1 mol O₂ / 31.9988 g O₂)

= 0.7813 mol O₂

Use the moles of O₂ and the mole ratio from the balanced chemical reaction and determine the number of moles of Al₂O₃ that will be produced as follows:

= 0.7813 mol O₂ x (2 mol Al₂O₃ / 3 mol O₂)

= 0.5209 mol Al₂O₃

Convert the moles to Al₂O₃ to grams as follows:

The molar mass of Al₂O₃ = 101.96 g/mol

= 0.5209 mol Al₂O₃ x ( 101.96 g Al₂O₃ / 1 mol Al₂O₃)

= 53.1 g Al₂O₃

Thus, 53.1 g of the product Al₂O₃ could be produced.