Question

Determine the stoichiometric amount of O2 necessary to react with 0.212 mol Al according to the following equation: 4Al (s) + 3 O2 (g) → 2 Al2O3 (s)

Answer 1

The stoichiometric amount of O₂ necessary to react with 0.212 mol Al is 0.159 mol O₂.

Step-by-step explanation:

In order to determine the stoichiometric amount of O₂ necessary to react with 0.212 mol Al according to the given equation, we need to use the mole ratios in the balanced equation.

According to the balanced equation, 4 moles of Al react with 3 moles of O₂ to produce 2 moles of Al2O3. Therefore, the stoichiometric ratio of Al to O₂is 4:3.

To find the stoichiometric amount of O₂, we can set up a proportion:

(0.212 mol Al) / (4 mol Al) = (x mol O₂) / (3 mol O₂)

Solving for x, we get:

x = (0.212 mol Al) * (3 mol O₂) / (4 mol Al) = 0.159 mol O₂

Therefore, the stoichiometric amount of O₂necessary to react with 0.212 mol Al is 0.159 mol O₂.