Chemistry 1
Volume 2
Worksheet 10
Limiting Reactants – Part 1
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1. Use the balanced reaction below to answer the following questions.
2 BF3 + 3 H2 → 2 B + 6 HF
a. If 0.10 mol of BF3 is reacted with 0.25 mol H2, which reactant is the limiting reactant?
b. What is the maximum amount (in grams) of HF that can be produced from these
amounts?
2. How many moles of NaCl are produced from the reaction between 20.0 g CuCl2 and 15.0 g
NaNO3, as outlined in the reaction below?
CuCl2 + 2 NaNO3 à Cu(NO3)2 + 2 NaCl
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3. The reaction for the combustion of ethane is shown by the balanced reaction below:
2 C2H6 + 7 O2 à 6 H2O + 4 CO2
1.45 g C2H6 was burned in the presence of 4.50 g O2. Determine the limiting reagent.
4. Aluminum sulfite reacts with sodium hydroxide as shown in the reaction below.
Al2(SO3)3 + 6 NaOH à 3 Na2SO3 + 2 Al(OH)3
If 5.0 g Al2(SO3)3 reacts with 4.2 g NaOH, how many moles of Al(OH)3 are produced?
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5. Which chemical is the limiting reactant in the following reaction if 5.0 g Zn are heated with
2.4 g MoO3?
3 Zn + 2 MoO3 à Mo2O3 + 3 ZnO
6. In the reaction below, 2.34 g I2O5 react with 3.40 g CO. How many grams of CO2 are
produced?
I2O5 + 5 CO à 5 CO2 + I2
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7. Propane (C3H8) is combusted via the reaction below. If 3.4 g propane are burned in the
presence of 3.4 g O2, how much H2O is produced?
C3H8 + 5 O2 à 3 CO2 + 4 H2O
8. Use the balanced equation below to answer the question that follows it.
8 CO + 17 H2 à C8H18 + 8 H2O
If 6.71 g CO and 8.00 g H2 are reacted, which is the limiting reagent?
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9. In the presence of HCl, aluminum metal produces aluminum chloride, as shown in the
reaction below.
2 Al + 6 HCl à 2 AlCl3 + 3 H2
If 2.67 g Al react with 3.5 g HCl, how many moles of AlCl3 are produced?
10. Consider the following equation to answer the question that follows:
2 As + 6 NaOH à 2 Na3AsO3 + 3 H2
If 9.43 g As reacts with 2.6 g NaOH, what is the maximum number of grams of H2 that
can be produced?
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Answer Key
1. Use the balanced reaction below to answer the following questions.
2 BF3 + 3 H2 → 2 B + 6 HF
a. If 0.10 mol of BF3 is reacted with 0.25 mol H2, which reactant is the limiting reactant?
To solve this problem, determine which of the two reactants produces the least
amount of product. Do this by converting the number of moles of each reactant
into moles of product. Choose either of the two products (B or HF) to convert
into, so long as the same product is used for both calculations. Use HF since the
next part of the question asks about this product.
0.10 mol BF3
6 mol HF
2 mol BF3
= 0.30 mol HF
0.25 mol H2
6 mol HF
3 mol H2
= 0.50 mol HF
Just look at which reactant produces the smallest number of moles of product.
In this case, BF3 produces a smaller amount of HF, so it is the limiting reactant.
Correct answer: BF3
b. What is the maximum amount (in grams) of HF that can be produced from these
amounts?
To do this, we look at the amount in moles of the limiting reactant in part a (0.30
mol HF). We then use this number to convert moles of HF to grams of HF using
the molar mass of HF (20.01 g/mol).
0.30 mol HF
20.01 g HF
1 mol HF
= 6.0 g HF
Correct answer: 6.0 g HF
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2. How many moles of NaCl are produced from the reaction between 20.0 g CuCl2 and 15.0 g
NaNO3, as outlined in the reaction below?
CuCl2 + 2 NaNO3 à Cu(NO3)2 + 2 NaCl
To solve this problem, we need to convert the grams of each reactant into moles of
reactant. Then, we need to convert those moles into moles of NaCl. We will then
compare which reactant produces the smallest number of moles of NaCl.
20.0 g CuCl2
1 mol CuCl2
134.45 g CuCl2
2 mol NaCl
1 mol CuCl2
= 0.298 mol NaCl
15.0 g NaNO3
1 mol NaNO3
84.995 g NaNO3
2 mol NaCl
2 mol NaNO3
= 0.176 mol NaCl
The maximum number of moles NaCl is 0.176 mol, since it is the lesser of the two
amounts produced from the reagents.
Correct answer: 0.176 mol NaCl
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3. The balanced reaction for the combustion of ethane is shown below:
2 C2H6 + 7 O2 à 6 H2O + 4 CO2
1.45 g C2H6 was burned in the presence of 4.50 g O2. Determine the limiting reagent.
Here, we once again need to convert the masses of reactants into moles of one of
the two products. Whichever one produces the least amount of the product is the
limiting reactant.
1.45 g C2H6
1 mol C2H6
30.07 g C2H6
6 mol H2O
2 mol C2H6
= 0.145 mol H2O
4.50 g O2
1 mol O2
31.998 g O2
6 mol H2O
7 mol O2
= 0.121 mol H2O
Since O2 produces fewer moles of H2O, it is the limiting reactant.
Correct answer: O2 is the limiting reactant.
4. Aluminum sulfite reacts with sodium hydroxide as shown in the reaction below.
Al2(SO3)3 + 6 NaOH à 3 Na2SO3 + 2 Al(OH)3
If 5.0 g Al2(SO3)3 reacts with 4.2 g NaOH, how many moles of Al(OH)3 are produced?
We need to do two things: determine the limiting reactant and use that information to
determine the number of moles of Al(OH)3 produced. The easiest way to do this is to
combine both calculations into one step, where we convert the amount of each reagent
into moles of the product. Whichever one produces the lower number of moles is the
limiting reagent, and the amount calculated is the amount of moles Al(OH)3 produced.
5.0 g Al2(SO3)3
1 mol Al2(SO3)3
294.15 g Al2(SO3)3
2 mol Al(OH)3
1 mol Al2(SO3)3
= 0.034 mol Al(OH)3
4.5 g NaOH
1 mol NaOH
39.997 g NaOH
2 mol Al(OH)3
6 mol NaOH
= 0.038 mol Al(OH)3
Al2(SO3)3 produces fewer moles of Al(OH)3, so it is the limiting reactant and 0.034 mol
Al(OH)3 is our answer.
Correct answer: 0.034 mol Al(OH)3 will be produced
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5. Which chemical is the limiting reactant in the following reaction if 5.0 g Zn are heated with
2.4 g MoO3?
3 Zn + 2 MoO3 à Mo2O3 + 3 ZnO
5.0 g Zn
1 mol Zn
65.38 g Zn
3 mol ZnO
3 mol Zn
= 0.076 mol ZnO
2.4 g MoO3
1 mol MoO3
143.94 g MoO3
3 mol ZnO
2 mol MoO3
= 0.025 mol ZnO
The limiting reactant is MoO3 since it produced fewer moles of ZnO.
Correct answer: MoO3 is the limiting reactant.
6. In the reaction below, 2.34 g I2O5 react with 3.40 g CO. How many grams of CO2 are
produced?
I2O5 + 5 CO à 5 CO2 + I2
2.34 g I2O5
1 mol I2O5
333.80 g I2O5
5 mol CO2
1 mol I2O5
44.01 g CO2
1 mol CO2
= 1.54 g CO2
3.40 g CO
1 mol CO
28.01 g CO
5 mol CO2
5 mol CO
44.01 g CO2
1 mol CO2
= 5.34 g CO2
Since I2O5 produces a lower amount of CO2, the amount of CO2 produced is 1.54 g.
Correct answer: 1.54 g CO2
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7. Propane (C3H8) is combusted via the reaction below. If 3.4 g propane are burned in the
presence of 3.4 g O2, how much H2O is produced?
C3H8 + 5 O2 à 3 CO2 + 4 H2O
3.4 g C3H8
1 mol C3H8
44.10 g C3H8
4 mol H2O
1 mol C3H8
18.02 g H2O
1 mol H2O
= 5.6 g H2O
3.4 g O2
1 mol O2
31.998 g O2
4 mol H2O
5 mol O2
18.02 g H2O
1 mol H2O
= 1.5 g H2O
Correct answer: 1.5 g H2O
8. Use the balanced equation below to answer the question that follows it.
8 CO + 17 H2 à C8H18 + 8 H2O
If 6.71 g CO and 8.00 g H2 are reacted, which is the limiting reagent?
6.71 g CO
1 mol CO
28.01 g CO
8.00 g H2
1 mol H2
2.02 g H2
8 mol H2O
8 mol CO
8 mol H2O
17 mol H2
= 0.240 mol H2O
= 1.86 mol H2O
CO produces the smallest number of moles of H2O, so it is the limiting reactant.
Correct answer: CO is the limiting reagent
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9. In the presence of HCl, aluminum metal produces aluminum chloride, as shown in the
reaction below.
2 Al + 6 HCl à 2 AlCl3 + 3 H2
If 2.67 g Al react with 3.5 g HCl, how many moles of AlCl3 are produced?
2.67 g Al
1 mol Al
26.98 g Al
2 mol AlCl3
2 mol Al
= 0.0990 mol AlCl3
3.5 g HCl
1 mol HCl
36.46 g HCl
2 mol AlCl3
6 mol HCl
= 0.032 mol AlCl3
Since HCl produces fewer moles of AlCl3, it is the limiting reagent, and the number of
moles of AlCl3 produced is 0.032 moles.
Correct answer: 0.032 mol AlCl3
10. Consider the following equation to answer the question that follows:
2 As + 6 NaOH à 2 Na3AsO3 + 3 H2
If 9.43 g As reacts with 2.6 g NaOH, what is the maximum number of grams of H2 that
can be produced?
9.43 g As
1 mol As
74.92 g As
3 mol H2
2 mol As
2.02 g H2
1 mol H2
2.6 g NaOH
1 mol NaOH
3 mol H2
39.997 g NaOH 6 mol NaOH
= 0.381 g H2
2.02 g H2
1 mol H2
= 0.066 g H2
NaOH produces the smaller number of grams of H2, so 0.066 g is the amount formed.
Correct answer: 0.066 g H2
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