LIMITING REAGENTS:

These are more than just ‘what you have less of’ in terms of moles or mass. You need to compare the ratios know what reagent will limit your reaction.

[EXAMPLE 1]

In the equation below what would be the limiting reagent?

(4g) H2 + (38g)F2 → (2g) HF

First, convert to moles:

M H2= [4g mol^-1]/[(1.01)x 2 mol] = 2 mol

M F2= [38g mol^-1]/[(19)x2 mol] = 1 mol

Now compare how many moles you need for the balanced equation and how many you have:

Required ratio= 1 H2 moles:1 F2 mole

Real ratio= 2 H2 moles:1 F2 mole

There fore because you dont have the required amount of F2, that would be the limiting agent. You would be left with approximately a mole of extra H2 in the case of an ideal reaction that reaches its end.

[EXAMPLE 2]

In the equation below what would the limiting reagent be given that you have 4g of Hydrogen and 28g of Nitrogen?

3 H2 + N2 → 2NH3

First convert g into moles:

H2 → [4g/mol^-1]/[1.01×2]= 2 moles

N2 → [28g/mol^-1]/[14×2] = 1 mol

Now compare ratios!

The equation ratio= 3:1

Real ratio= 2:1

Here Hydrogen would be the limiting agent because in order for the mole of Nitrogen to be completely reacted the reaction requires another mole of Hydrogen.