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)
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
What is the amount (in moles) of excess reactant remaining if all of the limiting reactant...
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