LIMITING REAGENT
Coach Cox
WHAT IS A LIMITING REAGENT?
• Limiting Reagent (Reactant) – the reactant in a chemical
reaction that limits the amount of product that can be formed.
The reaction will stop when all of the limiting reactant is
consumed
• Excess Reactant – the reactant in a chemical reaction that
remains when a reaction stops when the limiting reactant is
completely consumed. The excess reactant remains because
there is nothing with which it can react.
EXAMPLE
• No matter how many tires there are, if there are only 8 car bodies, then only
8 cars can be made. Likewise with chemistry, if there is only a certain
amount of one reactant available for a reaction, the reaction must stop when
that reactant is consumed whether or not the other reactant has been used
up.
EXAMPLE LIMITING REAGENT CALCULATION
A 2.00 g sample of ammonia is mixed with 4.00 g of
oxygen. Which is the limiting reactant and how much excess
reactant remains after the reaction has stopped?
First, we need to create a balanced equation for the reaction:
___ NH3(g) + ___ O2(g)  ___ NO(g) + ___ H2O(g)
EXAMPLE LIMITING REAGENT CALCULATION
• Next we can use stoichiometry to calculate how much product is
produced by each reactant.
• NOTE: It does not matter which product is chosen, but the
same product must be used for both reactants so that the
amounts can be compared.
EXAMPLE LIMITING REAGENT CALCULATION
• Balanced Equation: 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
•
Given in | Convert to | Mole to
| Convert to
Question | Moles (MM) | Mole Ratio | grams (MM)
2.00g NH3 | 1 mol NH3 | 4 mol NO
| 30.0 g NO
= 3.53 g NO (excess - EX)
| 17.0 g NH3 | 4 mol NH3 | 1 mol NO
•
4.00 g O2 | 1 mol O2 | 4 mol NO | 30.0 g NO
| 32.0 g O2 | 5 mol O2 | 1 mol NO
=
3.00 g NO (limiting - LR)
EXAMPLE: LIMITING REAGENT
CALCULATION ANSWER
• The reactant that produces the least amount of product
in this case is the oxygen, which is the "limiting
reactant."
EXAMPLE: LIMITING REAGENT
CALCULATION ANSWER
• Next, to find the amount of excess reactant, we must calculate
how much of the non-limiting reactant (ammonia) actually did
react with the limiting reactant (oxygen).
•
•
4.00 g O 2 | 1 mol O2 | 4 mol NH3 | 17.0 g NH 3
| 32.0 g O 2 | 5 mol O2
| 1 mol NH3
= 1.70 g NH3
EXAMPLE: LIMITING REAGENT
CALCULATION ANSWER
• We're not finished yet though. 1.70 g is the amount of ammonia
that reacted, not what is left over. To find the amount of excess
reactant remaining, subtract the amount that reacted from the
amount in the original sample.
• 2.00 g NH3 (original sample) – 1.70 g (reacted) = 0.30 g NH3 remaining