Limiting Reagent

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esob

Article 14
Staff member
7+ Year Member
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This showed up as the MCAT question of the day but I think it's wrong and wanted some confirmation. Shouldn't 64 g of oxygen be divided by the weight of diatomic oxygen (32) and thus yield on 2 moles of diatomic oxygen. This then should be oxygen, not ammonia is the limiting reagent

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Hi @esob -

Couple of things going on here. First, you're right that you need to use diatomic oxygen, so you have two moles of O2. However, there's a twist: the equation they give actually isn't balanced. The balanced equation is 4 NH3 + 5 O2 = 4 NO + 6 H2O. Since you have 3 moles of NH3 and 2 moles of O2, it will actually still be the case that oxygen is the limiting reagent.

Good catch though!

Hi @esob -

Couple of things going on here. First, you're right that you need to use diatomic oxygen, so you have two moles of O2. However, there's a twist: the equation they give actually isn't balanced. The balanced equation is 4 NH3 + 5 O2 = 4 NO + 6 H2O. Since you have 3 moles of NH3 and 2 moles of O2, it will actually still be the case that oxygen is the limiting reagent.

Good catch though!

right, I factored the balanced equation in when determining oxygen as the limiting reagent. It's actually the standard Khan academy video formula, except in their video they only have 32 grams.

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