What is the excess reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams
Of O₂?
Reaction: 4Al + 3O₂ → 2Al₂O₃

Question

What is the excess reactant for 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

To determine the excess reactant, we need to calculate how much of each reactant is actually consumed in the reaction. We can do this by calculating how much of each reactant is required to react completely with the other reactant, and then comparing that to the actual amount of each reactant that we have. Starting with 10.0 grams of Al and 19.0 grams of O₂, we can use their molar masses to calculate the number of moles of each: moles of Al = 10.0 g / 26.98 g/mol = 0.371 mol moles of O₂ = 19.0 g / 281.16 g/mol = 0.0676 mol We can see that Al is in excess, as we have more moles of Al than we need to react completely with the O₂. To calculate how much of the O₂ is needed to react completely with the Al, we can use the balanced equation to determine the mole ratio: 4Al + 3O₂ → 2Al₂O₃ 1 mol of Al reacts with 3/4 mol of O₂, so to react completely with the 0.371 mol of Al we have, we need: mol of O₂ = 0.371 mol Al x (3/4 mol O₂ / 1 mol Al) = 0.278 mol O₂ This is less than the 0.0676 mol of O₂ we actually have, so O₂ is the limiting reactant and Al is the excess reactant. Therefore, the answer is B) O₂.

What Is the Excess Reactant for the Reaction Below Given