Limiting reactants:
First off, you have to be able to identify a limiting reactant problem in the first place, otherwise you are likely do too
much (or too little) work. As I said in class, you will know that you are doing a limiting reactant problem when you are
given an amount of more than one reactant and asked to calculate how much of a product has been formed. You will have
to do one calculation for each amount of starting material given!
Example:
What mass of PbCl2 will be made when 15.9 grams of CaCl2 is added to a solution of Pb(NO3)2?
This is NOT a limiting reactant problem because you are given an amount of only a single
reactant. You need only do one calculation.
Example:
What mass of Ti(CO3)2 will be made when 0.558 grams of Ti(C2H3O2)4 is added to a solution containing
1.13 grams of Rb2CO3?
This IS a limiting reactant problem because you are given an amount of more than one reactant
(Ti(C2H3O2)4 and Rb2CO3) and asked for an amount of product (Ti(CO3)2) formed. You will
have to do two separate calculations!
If the question is NOT a limiting reactant problem, you simply convert the amount of the stuff that you were given to the
amount of the stuff you asked for. (What do you ALWAYS have to use when going from an amount of one substance to
an amount of another substance?) If it IS a limiting reactant questions, it is the exact same calculation, you just do it more
than once!
In a limiting reactant problem, you have set amounts of reactants, one of which you are likely to run out of while some of
the other reactant remains (think about the tricycle example I did in class).
Example:
Hydrogen gas and oxygen gas are mixed to make water. If 2.80 grams of hydrogen gas is reacted with
17.7 grams of oxygen gas, how much water can be formed and how much of the reactants will be left
over?
The first step, as usual, is to get a balanced equation (you must do this for any of these problems), so:
2H2(g) + O2(g)  2H2O(l)
The second step is to see how much of a product you can make from each given amount of reactant.
2.80 g H 2 
1 mol H 2 2 mol H 2O 18.02 g H 2O


 25.0 g H 2O
2.02 g H 2 2 mol H 2
1 mol H 2O
17.8 g O2 
1 mol O2 2 mol H 2O 18.02 g H 2O


 20.0 g H 2O
32.00 g O2 1 mol O2
1 mol H 2O
So, you can use up all of your H2 and make 25.0g of H2O or you can use up all of your O2 and make 20.0
g of H2O. You only have 17.8 g of O2, so you cannot make any more H2O than the 20.0 g shown
in the second calculation. You ran out of O2 before you ran out H2, so the O2 was the limiting
reactant.
From above, we have found that you can make 20.0 g of H2O (O2 was limiting) from the given amounts
of H2 and O2. This means that the theoretical yield of H2O is 20.0 g. The next part of the
question is how much of each reactant is left after the reaction is complete.
Finding how much of the limiting reactant is left is a simple as asking yourself “How much of something
is left when I run out of it?” The answer? None! By definition, if you run out of something, you
have none of it left which means that the remaining amount of limiting reactant is always zero!
There are two ways to find out how much of the excess reactant is left. In one, you must first find out
how much of the excess reactant was used in the reaction. You will do this by starting with the
amount of the limiting reactant you were given in the question (17.8 g of O2) and converting it to
the excess reactant (H2)
17.8 g O2 
1 mol O2 2 mol H 2 2.02 g H 2


 2.24 g H 2
32.00 g O2 1 mol O2 1 mol H 2
This is the amount of the H2 that was USED to make the H2O!
2.24 g H2 is not what is left over, it is what you USED. You MUST subtract this amount from
what you started with to find out how much is LEFT!
2.80 g H2 (started) – 2.24 H2 (used) = 0.56 g H2 LEFT OVER AFTER THE REACTION!
The second way (you can choose your favorite approach since they will give virtually the same
answer) is to remember the “Law of Conservation of Mass”. You must start and end with the
same mass! In other words, the total mass of reactants you had before the reaction started must
be equal to the total mass of all products AND remaining reactants after the reaction is complete.
17.7 g O2 + 2.80 g H2 = 20.5 g total reactants before the reaction
20.0 g H2O + 0 g O2 + X g H2 = 20.5 g total after the reaction
X = 0.5 g H2 left after the reaction!
The difference between the two methods (0.56 g vs. 0.5 g) is due solely to sig. figs.
This was fairly long-winded, but hopefully helpful. The next part will be a more condensed version of “How to do
limiting reactant problems”
Once you have identified a problem as a limiting reactant problem, here are the steps to go through:
Step 1: Write the balanced chemical equation
Step 2: Pick a product and find out what mass of that product can be made from each of the given reactants.
YOU MUST ALWAYS COMPARE THE AMOUNT THE SAME PRODUCT MADE FROM EACH
REACTANT. If the reaction is A + B  C + D + E, then you must pick ONE product, let’s say E in the
case, and find out how much E can be made from A and how much E can be made from B. The
smallest of the calculated product masses is how much of the product can actually be made (this is the
theoretical yield)! The reactant leading to the smallest amount of product is the limiting reactant.
Remember, all of the limiting reactant is used up, so the amount remaining is ALWAYS zero.
The following only needs to be done if the question asks for masses of the other products and reactants
Step 3: For the rest of the calculations, start with the amount of the limiting reactant given in the question.
Calculate the masses of other products made from the given amount of the limiting reactant (since these
are products, they are the amount MADE and therefore no subtraction is needed).
Step 4: Calculate the masses of the excess reagent(s) USED starting with the amount of the amount of the limiting
reagent given. Reagents are always USED, so once you have calculated the amount used, you MUST
subtract it from the amount of that reactant you were given.
I recommend making a table like I showed in class to keep track of everything!
Example 1: 4.529 g of antimony (III) sulfate and 13.98 g of lead (II) chlorate are reacted together. How much of
all products and reactants are left after the reaction is complete?
This IS a limiting reactant question because amounts of the two reactants are given.
Step 1: balance equation:
Sb2(SO4)3(aq) + 3 Pb(ClO3)2(aq)  2 Sb(ClO3)3(aq) + 3 PbSO4(s)
Step 2: I am going to pick the lead (II) sulfate as my product just because I can. It doesn’t matter what
product you start with, you will get the same answers!
1 mol Sb 2 (SO 4 )3
3 mol PbSO 4
303.3 g PbSO 4


 7.750 g PbSO 4
531.71 g Sb 2 (SO 4 )3 1 mol Sb 2 (SO 4 )3 1 mol PbSO 4
1 mol Pb(ClO 3 )2
3 mol PbSO4
303.3 g PbSO 4
13.98 g Pb(ClO 3 )2 


 11.33 g PbSO 4
374.1 g Pb(ClO 3 )2 3 mol Pb(ClO 3 )2 1 mol PbSO 4
4.529 g Sb 2 (SO 4 )3 
7.750 g PbSO4 is the smallest amount of product calculated, so that is how much can actually be
made. This also shows that Sb2(SO4)3 is the limiting reactant (of which there will be none left).
Sb2(SO4)3 = 0 g
Pb(ClO3)2
Sb(ClO3)3
Pb(SO4)2 = 7.750 g
Step 3:
1 mol Sb 2 (SO 4 )3
2 mol Sb(ClO 3 )3 372.10 g Sb(ClO 3 )3


 6.339 g Sb(ClO 3 )3
531.68 g Sb 2 (SO 4 )3 1 mol Sb 2 (SO 4 )3
1 m ol Sb(ClO 3 )3
This is the amount of the second product that can actually be made!
4.529 g Sb 2 (SO4 )3 
Sb2(SO4)3 = 0 g
Pb(ClO3)2
Sb(ClO3)3 = 6.339 g
Pb(SO4)2 = 7.750 g
Step 4:
4.529 g Sb 2 (SO 4 )3 
1 mol Sb 2 (SO 4 )3
3 mol Pb(ClO 3 )2 374.09 g Pb(ClO 3 )2


 9.560 g Pb(ClO 3 )2
531.68 g Sb 2 (SO 4 )3 1 mol Sb 2 (SO 4 )3
1 m ol Pb(ClO 3 )2
This is the amount of the excess reactant that was USED!!
13.98 g Pb(ClO3)2 (started) – 9.560 g Pb(ClO3)2 (used) = 4.42 g Pb(ClO3)2 LEFT!!
This last step is the step the most people forget, so DON’T forget!
OR
(started masses) – (final masses) = excess reactant left
(4.529 + 13.98) – (6.339 + 7.750) = 4.42 g excess left
Either way, the final answer is:
Sb2(SO4)3 = 0 g
Pb(ClO3)2 = 4.42 g
Sb(ClO3)3 = 6.339 g
Pb(SO4)2 = 7.750 g
Example 2: What is the mass of BOTH products after the reaction between 1.14g of potassium carbonate with
1.15g of tin (IV) nitrate?
2 K2CO3(aq) + Sn(NO3)4(aq)  Sn(CO3)2(s) + 4 KNO3(aq)
1 mol K 2CO3
1 mol Sn(CO3 )2 238.71 g Sn(CO3 )2


 0.984 g Sn(CO3 )2
138.213 g K 2CO3 2 mol K 2CO3
1 mol Sn(CO3 )2
1 mol Sn(NO3 )4
1 mol Sn(CO3 )2 238.71 g Sn(CO3 )2
1.15 g Sn(NO3 )4 


 0.815 g Sn(CO3 )2
336.75 g Sn(NO3 )4 1 mol Sn(NO3 )4
1 mol Sn(CO3 )2
1.14 g K 2CO3 
0.815 g Sn(CO3)2 can be made and Sn(NO3)4 is the L.R (none left).
Once you have run out of one of the reactants, you CANNOT make ANY more of ANY of the
products. This means that once you have identified the limiting reactant for one product, the same
reactant will be the limiting reactant for ALL PRODUCTS!!! The only reason you had to do 2
calculations (one for each of the reactants) above was because you had to identify the limiting
reactant. Now that you know what the L.R. is, you only need to do one calculation for each
remaining products starting with the given amount of the L.R.
1.15 g Sn(NO3 )4 
1 mol Sn(NO3 )4
4 mol KNO3
101.11 g KNO3


 1.38 g KNO3
336.75 g Sn(NO3 )4 1 mol Sn(NO3 )4 1 mol KNO3
Sn(CO3)2 = 0.814 g made
KNO3 = 1.38 g made
You try: (Calculate the masses of ALL reactants and the theoretical yield of all products after the given reactions take
place)
1)
25.00 g of nickel (II) chloride is reacted with 25.00 g of silver sulfate
2)
18.9 grams of lithium phosphate reacts with gold (I) acetate
3)
A vessel contains 0.611 grams of butane gas (C4H10(g)) and 1.358 liters of oxygen gas (at STP). After the butane
is reacted with the oxygen, what is the theoretical yield (in mL) of CO2(g) (at STP) and theoretical yield (in
molecules) of water?
4)
0.500 grams of copper (II) nitrate is reacted with 1.25 grams of sodium oxalate
5)
525 mL of 1.199 M sulfuric acid is added to 23.66 grams of strontium bromide
6)
3.8 grams of octane (C8H18) is burned in 22.1 grams of oxygen gas.
7)
10.0 mL of 1.50 M solution of hydroiodic acid is poured onto 0.95 g of solid calcium carbonate.
2 HI(aq) + CaCO3(s)  CaI2(aq) + H2O(l) + CO2(g)