Question

What is the amount (in moles) of excess reactant remaining if all of the limiting reactant completely reacts when 24.9 mol of aluminum and 27.9 moles of oxygen gas react?

4 Al (s) + 3 O₂(g) → 2 Al₂O₃(s)

Answer 1

Balanced chemical equation is:

4 Al + 3 O2 ---> 2 Al2O3 +

4 mol of Al reacts with 3 mol of O2

for 24.9 mol of Al, 18.67 mol of O2 is required

But we have 27.9 mol of O2

so, Al is limiting reagent

we will use Al in further calculation

According to balanced equation

mol of O2 reacted = (3/4)* moles of Al

= (3/4)*24.9

= 18.67 mol

mol of O2 remaining = mol initially present - mol reacted

mol of O2 remaining = 27.9 - 18.67

mol of O2 remaining = 9.225 mol

Answer: 9.22 mol