Sciencemadness Discussion Board

determining the limiting reagent?

agentsalsa - 31-1-2006 at 15:28

i have a chem final tomorrow and i forgot how to determine the limiting reagent. help would be greatly appreciated! thanks :)

Magpie - 31-1-2006 at 16:02

1. Write a balanced equation for the reaction.
2. Determine the g-moles of reactants available. Do this by dividing the weight in grams available by the molecular weight for each reactant.
3. The coefficients of the reactants from the balanced equation tell you the mole ratio(s) required.
4. Compare these ratio(s) to the ratio(s) of the g-moles available as determined in step 2. You will see which reactant is short(est) in g- moles. This is the limiting reactant.

Say you had

2H2 + O2 -> 2H2O and your starting quantities were 5 grams of H2 and 40 g of O2. Which is the limiting reactant? Try it and let us know. ;)

agentsalsa - 31-1-2006 at 16:23

okay, so would O2 be the limiting reactant?

i did 5 divided by the gram formula mass of H2(2.02) and 40 divided by the gram formula mass of O2(32)

5 divided by 2.02=2.475

40 divided by 32=1.25

so is O2 the limiting reactant?

Darkblade48 - 31-1-2006 at 19:35

Quote:
Originally posted by agentsalsa
okay, so would O2 be the limiting reactant?

i did 5 divided by the gram formula mass of H2(2.02) and 40 divided by the gram formula mass of O2(32)

5 divided by 2.02=2.475

40 divided by 32=1.25

so is O2 the limiting reactant?

No, you cannot simply look at the number and determine which one is smaller, you have to remember to look at the stochiometric ratio.

In this case, for every 1 mole of oxygen, you require 2 moles of hydrogen. So, 1.25 moles of oxygen would require 2.5 moles of hydrogen, which you do not have; therefore hydrogen is the limiting reagent.

Here's another practice problem:

Propane (C3H8) + O2 --> CO2 + H2O

You start with 4.8 grams of propane and 12.3 grams of oxygen. Which is the limiting reagent?

Magpie - 31-1-2006 at 19:45

My random choice of weights, it turns out was very poor as an illustration, and you shall see why:

Using your numbers the mole ratio of H2/O2 available is

2.475/1.25 = 1.98

Then from the equation coefficients

moles H2/moles O2 = 2/1 = 2

So technically, H2 is the limiting reagent as we are slightly short of H2. Hope this helps.