How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂?
Reaction: 4Al + 3O₂ → 2Al₂O₃

Question

How many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O₂? Reaction: 4Al + 3O₂ → 2Al₂O₃

Quizplus · Accepted Answer

First, we need to determine the limiting reactant to see how much product can be formed. To do this, we need to convert the masses of Al and O₂ to moles: moles of Al: 10.0 g Al / 26.98 g/mol Al = 0.371 mol Al moles of O₂: 19.0 g O₂ / 48.00 g/mol O₂ = 0.396 mol O₂ Next, we need to determine the mole ratio of Al to O₂ from the balanced equation: 4Al + 3O₂ → 2Al₂O₃ 4 moles Al / 3 moles O₂ Now we can calculate the limiting reactant: 0.371 mol Al * (3 mol O₂ / 4 mol Al) = 0.278 mol O₂ Since 0.278 mol O₂ is less than the initial amount of 0.396 mol O₂, O₂ is the limiting reactant. We can now use the mole ratio from the balanced equation to determine the amount of Al₂O₃ formed: 0.278 mol O₂ * (2 mol Al₂O₃ / 3 mol O₂) = 0.185 mol Al₂O₃ Finally, we can convert mol Al₂O₃ to grams: 0.185 mol Al₂O₃ * 150.00 g/mol Al₂O₃ = 27.75 g Al₂O₃ Rounding to two significant figures, the answer is D) 18.9.

How Many Grams of Aluminum Oxide Are Produced According to the Reaction