Question

Consider the following reaction.

4Al(s) + 3O₂(g) → 2Al₂O₃(s)
It takes 9.00 L pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Answer 1

Using stoichiometry, we find that 14.47 grams of aluminum reacted with 9.00 L of oxygen at STP based on the molar volumes of gases and the balanced chemical equation.

Step-by-step explanation:

To calculate the mass of aluminum reacted with 9.00 L of pure oxygen gas at standard temperature and pressure (STP), we look at the stoichiometry of the balanced equation: 4Al(s) + 3O₂(g) → 2Al₂O₃(s).

At STP, 1 mole of any gas occupies 22.4 L. The molar ratio between oxygen and aluminum is 3:4. First, calculate the moles of O₂.

1. 9.00 L O₂ ÷ 22.4 L/mol = 0.402 moles O₂
2. Moles of Al reacted = 0.402 moles O₂ × (4 moles Al / 3 moles O₂) = 0.536 moles Al

Next, calculate the mass of aluminum using its molar mass (26.98 g/mol).

1. Mass of Al = 0.536 moles Al × 26.98 g/mol = 14.47 g Al

Therefore, the mass of aluminum that reacted with 9.00 L of oxygen at STP is 14.47 grams.