Limiting Stoichiometry Problems

What is a limiting Reactant?

The Limiting reactant runs out first and limits the amount of product that can be made.

What is the 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.

 

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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.
How to I determine which reactant is limiting?

To do this, for all reactants calculate the amount of product that would be produced if all of the reactant was used and there was excess of all the other reactants. The reactant that has the least product is the limiting reactant.

Example

Which is the limiting reactant and how much H2 is produced if there is 0.30 mol of Zn and 0.52 moles of HCl?

Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)

You must pick the same product for both calculations.

.30 moles of Zn makes .30 moles of H2 .52 moles of HCl makes .26 moles of H2

Since HCl produces the lesser amount it is the limiting reactant, the number of moles of H2 produced is .26 moles.

 

Mercury and bromine will react with each other to produce mercury(II) bromide.

Hg(l) + Br2(l) --> HgBr2(s)

(a) What mass of HgBr2 is produced from the reaction of 9.80 g Hg and 12.0 g Br2?
What mass of which reactant is left unreacted?


(b) What mass of HgBr2 is produced from the reaction of 5.50 mL of mercury (density = 13.6 g/mL) and 5.50 mL bromine (density = 3.10 g/mL)?

 
this is a limiting stoichiometry problem

you need a balanced equation                        Hg(l) + Br2(l) --> HgBr2(s)
convert both Hg and Br2 to moles

Which ever one is lesser is the limiting and is the one quantity you will use to calculate the product. You will have too much of the other and we call that excess.

9.80g Hg (1mol/200.59g)=0.0489mol Hg Limiting

12.0 Br2 (1mol/159.8g)=0.075mol Br2

Convert moles of Hg to moles og HgBr2 using the mole ratio from the balanced equtaion

0.0489mol Hg(1 mol HgBr2/1mol Hg)= 0.0489mol HgBr2

Convert the moles to HgBr2 to grams

0.0489mol HgBr2(360.3g/mol)=17.6g HgBr2

How much ractant is left?    17.6g product- 9.80 g Hg Used= 7.80g Br2 used

12.0g Br2 -7.80g Br2 used= 4.20g Left over

B. Hg 5.50mL(13.6g/mL)=74.8g Hg
74.8g Hg (1mol/200.59g)=0.373mol Hg 

Br2 5.50 mL (3.10/mL)=1.77g Br2
1.77g Br2(1mol/159.8g)=0.0111mol Br2 <--Limiting

0.0111mol Br2 (1 mol HgBr2/1mol Br2)= 0.0111mol HgBr2
0.0111mol HgBr2(360.3g/mol)=4.00g HgBr2