Limiting Reagent

[MedStudentWanna]

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Can someone explain how to do limiting reagent problems in terms of the MCAT? I thought I remembered from Chem, but apparently forgot because I missed every problem on it in the EK 1001 book.

 

[phospho]

Can someone explain how to do limiting reagent problems in terms of the MCAT? I thought I remembered from Chem, but apparently forgot because I missed every problem on it in the EK 1001 book.

what exactly is bothering you about those problems? identifying the limiting reagent? If so, let's say they give you the equation:

2A + 3B -------> 5C + 2D

You're given 4 grams of A and 5 grams of B, who is the limiting reagent, and why?

You convert A and B into moles (grams/MW=moles)

Let's say you end up getting 1 mole for A and 2 moles for B. We can do it like this:

put the value you got for A under A in the equation. Cross multiply to see how much B that would give you (1x3/2=3/2). You would need 3/2 moles of B for that one mole of A. Our calculation gave us 2 moles for B, so we have more than enough of B, right? Therefore, A is the limiting reagent.

You can also do it using B, and seeing whether you have enough A. You'll get the same answer.

You can also divide the # of moles of A you calculated by the coeffficient of A in the equation. Do the same with B. Compare the two values, and the smaller one is the limiting reagent.

Hopefully I didn't make this even more confusing. Best of luck

 

[Revilla]

Phospho explained it really well.

Limiting reagent problems are quick once you get used to doing them. If you're given grams, convert it to moles and then divide the number of moles by the coefficient (assuming you've balanced the equation) in front of the reactant.

If you're given moles, just divide by the coefficient in front of the reactants.

The smaller number is your limiting reagent.

 

[phospho]

what exactly is bothering you about those problems? identifying the limiting reagent? If so, let's say they give you the equation:

2A + 3B -------> 5C + 2D

You're given 4 grams of A and 5 grams of B, who is the limiting reagent, and why?

You convert A and B into moles (grams/MW=moles)

Let's say you end up getting 1 mole for A and 2 moles for B. We can do it like this:

put the value you got for A under A in the equation. Cross multiply to see how much B that would give you (1x3/2=3/2). You would need 3/2 moles of B for that one mole of A. Our calculation gave us 2 moles for B, so we have more than enough of B, right? Therefore, A is the limiting reagent.

You can also do it using B, and seeing whether you have enough A. You'll get the same answer.

You can also divide the # of moles of A you calculated by the coeffficient of A in the equation. Do the same with B. Compare the two values, and the smaller one is the limiting reagent.

Hopefully I didn't make this even more confusing. Best of luck

wow, as an attending now, what I said above is literally gibberish to me...