If you mix 12 mol H2 and 7 moles of O2 to form H2O, which is the limiting reagent and why?
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To solve, we must first write the balanced chemical equation:
__H2(g) + __O2(g) --> __H2O(l)
We have the same number of hydrogen on each side, but not of oxygen. To fix, we can place a 2 in front of the H2O molecule, giving us 2 oxygen on each side. This throws our hydrogen out of whack though. To correct, we now have 2 H on the left, and 4 on the right, so all we need to do is place a 2 before the H2(g) on the left to balance everything out.
From here, we see which of the given quantities (12 mol H2 or 7 mol O2) give us the lesser amount of water, as that will be the limiting reagent. This is a stoichiometry problem.
12 mol H2 x 2 mol H2O/2 mol H2 = 12 mol H2O
7 mol O2 x 2 mol H2O/1 mol O2 = 14 mol H2O
Since we can only produce 12 moles of H2O before the H2(g) is all used up, H2(g) is the limiting reagent.
(If it's still confusing for you, imagine being at a small school dance with 50 boys and 53 girls. If all 50 boys ask 50 girls to dance, there will still be three girls sitting out. Hence, the boys are the limiting reagent. Without more boys, there will be three excess girls who will be sitting out a dance. You can't have more partners on the dance floor than you have pairs)
__H2(g) + __O2(g) --> __H2O(l)
We have the same number of hydrogen on each side, but not of oxygen. To fix, we can place a 2 in front of the H2O molecule, giving us 2 oxygen on each side. This throws our hydrogen out of whack though. To correct, we now have 2 H on the left, and 4 on the right, so all we need to do is place a 2 before the H2(g) on the left to balance everything out.
From here, we see which of the given quantities (12 mol H2 or 7 mol O2) give us the lesser amount of water, as that will be the limiting reagent. This is a stoichiometry problem.
12 mol H2 x 2 mol H2O/2 mol H2 = 12 mol H2O
7 mol O2 x 2 mol H2O/1 mol O2 = 14 mol H2O
Since we can only produce 12 moles of H2O before the H2(g) is all used up, H2(g) is the limiting reagent.
(If it's still confusing for you, imagine being at a small school dance with 50 boys and 53 girls. If all 50 boys ask 50 girls to dance, there will still be three girls sitting out. Hence, the boys are the limiting reagent. Without more boys, there will be three excess girls who will be sitting out a dance. You can't have more partners on the dance floor than you have pairs)
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2H2 + O2------>2H2O
12 7 0
0 1
See the mole ratio.Thus H2 is limiting reagent.
12 7 0
0 1
See the mole ratio.Thus H2 is limiting reagent.