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CHEM I Handout – Limiting Reactant Practice
Limiting Reactant Practice
Sometimes when we are given a problem, we are given information about more than one of the
reactants. When this happens, we need to figure out which reactant will run out first! Take
making sandwiches. First, say we have a tablespoon of peanut butter and a whole loaf of bread,
how many sandwiches can be made? One sandwich can be made, right? Now, what if we have
a whole jar of peanut butter and six slices of bread, how many sandwiches can be made? This
time we can make three, right? Before we could figure out how many sandwiches could be made
in each case, we had to determine what would run out first. In chemical problems, we will have
to do the same thing!
1. To determine the limiting reactant in a problem, do we look at the reactants (the bread and
the peanut butter) or the products (the sandwiches)?
2. To determine the limiting reactant, will we use grams, moles, or number of particles?
Ready to try some calculations?
There are two approaches: 1) determine which reactant runs
out first using mole ratios and only do a calculation with it, or 2) calculate the final answer for
both reactants and the smaller answer is the correct answer. Remember to use the same problemsolving skills (roadmap and such) that we have already learned. The first problem below is
worked out using both methods. The key will only show method 2 for the other problems, but
you may use whichever method you prefer!
3. For the chemical reaction shown below, what is the maximum amount of zinc sulfide, in
grams, that can be made from 0.503 g of sulfur and 0.987 g of zinc.
8 Zn + S8 →
0.987 g
0.503 g
8 ZnS
?g
Method 1 Solution
0.987 g Zn
1 mol Zn
 0.0151 mol Zn 0.00196  7.70 mol Zn
65.38 g Zn
1 mol S8
0.503 g S8 
 0.00196 mol S8  0.00196  1.00 mol S8
256.52 g S8
Since we only have a 7.70:1.00 ratio and need an 8:1
ration, zinc limits. Thus, only need to calculate with
zinc:
0.0151 mol Zn
8 mol ZnS 97.45 g ZnS

 1.47 g ZnS
8 mol Zn
0.987 g Zn
1 mol Zn
Method 2 Solution

8 mol ZnS 97.45 g ZnS

 1.47 g ZnS
1 mol ZnS
97.45
g ZnS
8 mol ZnS 
8
 1.53 g ZnS
0.503 g S8 


1
mol
ZnS
256.52 g S8 1 mol S8
65.38 g Zn
1 mol S
8 mol Zn
Now both of these answers can’t be correct, so we pick
the smaller one as our final answer: 1.47 g ZnS. The
smaller answer was based on zinc, so zinc limits.
1 mol ZnS
Limiting Reactant Key.docx
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CHEM I Handout – Limiting Reactant Practice
4. What is the theoretical yield, in grams, of sulfur dioxide that can be made from 0.132 moles
iron(II) sulfide and 0.497 moles of oxygen?
4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)
0.132 mol
0.497 mol
?g
5. If 2.00 grams of magnesium reacts with 0.0300 moles of silver nitrate, how many grams of
silver metal will be produced?
2 AgNO3(aq) + Mg(s) → Mg(NO3)2(aq) + 2 Ag(s)
0.0300 mol
2.00 g
?g
6. What is the theoretical yield of dinitrogen tetrafluoride, in grams, if we start with 15.89
grams of both ammonia and fluorine?
2 NH3(g) + 5 F2(g) → N2F4(g) + 6 HF(g)
15.89 g
15.89 g
?g
7. What is the maximum amount of water that can be obtained from 50.00 g of copper and
75.00 g of nitric acid according to the below reaction?
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(ℓ)
50.00 g
75.00 g
?g
8. If 50.00 g each of copper(II) chloride and sodium phosphate are mixed, what is the
theorectical yield of copper(II) phosphate that can be made?
3 CuCl2 (aq) + 2 Na3PO4 (aq) → Cu3(PO4)2(s) + 6 NaCl(aq)
50.00 g
Limiting Reactant Key.docx
50.00 g
?g
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CHEM I Handout – Limiting Reactant Practice
What if we are asked how much of the excess reagent was left over? As before, we can
determine two ways: 1) use the limiting reactant to determine how much of the excess was
really needed, then do a subtraction between that and how much was given to see how much was
excess, or 2) use the theoretical yield of the product to determine how much of the excess was
really needed, then do a subtraction between that and how much was given to see how much was
excess. Method one is the safer approach in case a mathematical error was made in determining
the theoretical yield, so only it will be used on the key.
9. For the chemical reaction shown below, what is the maximum amount of zinc sulfide, in
grams, that can be made from 0.503 g of sulfur and 0.987 g of zinc. Now, what is the mass of
the left over excess reagent?
8 Zn + S8 →
0.987 g
0.503 g
?g leftover
8 ZnS
? g = 1.47 g with Zn limiting (determined in #3)
Method 1 Solution
0.987 g Zn
Method 2 Solution
1 mol Zn

1 mol S8
65.38 g Zn 8 mol Zn

256.52 g S8
 0.484 g S needed
8
1 mol S8
We were given 0.503 g S8, but can now see we only
needed 0.484 g S8, so the excess sulfur is:
0.503 total g S8 – 0.484 needed g S8 = 0.019 excess g S8
1.47 g ZnS
1 mol ZnS

1 mol S8  256.52 g S8  0.484 g S needed
97.45 g ZnS 8 mol ZnS
1 mol S8
We were given 0.503 g S8, but can now see we only
needed 0.484 g S8, so the excess sulfur is:
0.503 total g S8 – 0.484 needed g S8 = 0.019 excess g S8
10. What is the theoretical yield, in grams, of sulfur dioxide that can be made from 0.132 moles
iron(II) sulfide and 0.497 moles of oxygen? In addition, how many moles of the excess
reagent are left over?
4 FeS2(s) + 11 O2(g) → 2 Fe2O3(s) + 8 SO2(g)
0.132 mol
? g = 16.91 g
0.497 mol
? mol leftover
0.497 total mol O2 – 0.363 needed mol O2 = 0.134 excess mol O2
11. If 2.00 grams of magnesium reacts with 0.0300 moles of silver nitrate, how many grams of
silver metal will be produced? Then, what mass of the excess reagent will there still be?
2 AgNO3(aq) + Mg(s) → Mg(NO3)2(aq) + 2 Ag(s)
0.300 mol
? g = 3.24 g
2.00 g
? g leftover
12. What is the theoretical yield of dinitrogen tetrafluoride, in grams, and the amount of leftover
excess reagent, in grams, if we start with 15.89 grams of both ammonia and fluorine?
2 NH3(g) + 5 F2(g) → N2F4(g) + 6 HF(g)
15.89 g
15.89 g
? g = 8.699 g
? g leftover
Limiting Reactant Key.docx
8
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CHEM I Handout – Limiting Reactant Practice
13. What is the maximum amount of water that can be obtained from 50.00 g of copper and
75.00 g of nitric acid according to the below reaction? Also, how many grams of reactants
are left over?
3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(ℓ)
50.00 g
? g = 10.72 g
75.00 g
? g leftover
14. If 50.00 g each of copper(II) chloride and sodium phosphate are mixed, what is the
theorectical yield of copper(II) phosphate that can be made and how many grams of excess
reagent are left over?
3 CuCl2 (aq) + 2 Na3PO4 (aq) → Cu3(PO4)2(s) + 6 NaCl(aq)
50.00 g
50.00 g
? g = 47.18 g
? g needed
Final thoughts…. Remember that if a problem gives information about more than one of the
reactants, it will be a limiting reactant problem! If information is only given about one reactant
or it says the other reactants are in excess, it is not a limiting reactant problem.
Limiting Reactant Key.docx
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