3.3.3 Finding Limiting Reagents

Question 1

Finding Limiting Reagents

Answer 1

• When two or more reagents are combined in a chemical reaction, one of them is the limiting reagent and will run out before the others (unless there are stoichiometric amounts of each).
• Determining how much product can be formed from the balanced chemical equation and the amounts of reagents available is the basis for the limiting-reagent problem in chemistry.

Question 2

note

Answer 2

• When two or more reagents are combined in a chemical reaction, one of them is the limiting reagent and will run out before the others.
• In this example, it takes 2 oz. of gin, 8 oz. of tonic, and 1/8 of a lime to make one gin and tonic. This is the balanced reaction. With 6 oz. of gin, 32 oz. of tonic and one lime on hand, it will only be possible to make three gin and tonics before one ingredient (the gin) runs out. The ingredient that runs out is the limiting ingredient (reagent).
• The ingredients on hand might present in the exact proportions in which they are required (in stoichiometric amounts). In this case they will all be exhausted at the same time. There are two methods for determining the limiting reagent in a problem.
• In the first method, assume one reagent is limiting and calculate how much of the other is needed. Either reagent can be assumed to be limiting.
• The example assumes PCl5 is limiting. The number of moles of water (H2O) is calculated that will react with this amount of PCl5. To do this, use the information in the balanced equation that PCl5 and H2O react in a 1:4 ratio.
• In this example, 0.8 moles of PCl5 would react with a total of 3.2 moles of water if it could. But there are only 2.9 moles of water present. Thus, water is the limiting reagent in this example.
• If there had been more than 3.2mol of water, the PCl5 would have been the limiting reagent.

Question 3

note 2

Answer 3

• In the second method, determine the maximum amount of one of the products (any one will work) from each of the reactants.
• In this example, begin by determining the maximum amount of H3PO4 possible from 0.8 moles of PCl5.
• Because PCl5 is consumed and H3PO4 is produced in a 1:1 ratio in this reaction, there will be 0.8 moles of H3PO4 produced.
• In a similar manner, determine the amount of H3PO4 produced from the amount of water provided. Because water is consumed and H3PO4 produced in a 4:1 ratio, there will be 0.72 moles of H3PO4 produced. This is less than the amount produced from PCl5. The conclusion is that water is the limiting reagent in this problem.
• As expected, both methods lead to the same answer.

Question 4

Which statement provides the best analogy for limiting reactants?

Answer 4

At the grocery store, packages of hot dogs contain 10 hot dogs, while packages of buns contain 8 buns. Because there are fewer buns than hot dogs, the buns are the limiting “reactant.”

Question 5

In the reaction shown, how many grams of hydrogen gas are produced when 23.10 g aluminum are reacted with 68.8 g HCl?

2Al + 6HCl → 2AlCl3 + 3H2

Answer 5

1.90 g

Question 6

Lithium and nitrogen react to form Li3N, according to the following balanced chemical equation.

6Li + N2 → 2Li3N 

When 4.0 grams of lithium reacted with 2.5 grams of nitrogen, how much of the non-limiting reactant is left over after the reaction is complete?

Answer 6

0.28 g Li

Question 7

Determine the mass of excess reactant remaining when 20.30 grams of ZnS and are reacted with 36.8 grams of oxygen.

2ZnS + 3O2 → 2ZnO + 2SO2

Answer 7

26.8 g

Question 8

The Contraption Manufacturing Corporation uses one bolt, three washers, and two nuts when making each contraption. Their current inventory is 4,000 bolts, 12,000 washers, and 7,000 nuts. How many contraptions can be manufactured?

Answer 8

3,500

Question 9

In the reaction shown

2Al + 6HCl → 2AlCl3 + 3H2 

How many moles of hydrogen gas are produced from 4.8 mol of HCl and 2.2 mol of Al?

Answer 9

2.4 mol

Question 10

6.0 mol of calcium chloride are reacted with 5.0 mol of sodium phosphate. What is the limiting reagent?

3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4 )2(s) + 6 NaCl(aq)

Answer 10

CaCl2

Question 11

How many moles of sulfur dioxide are formed from the reaction of 20.30 grams of zinc sulfide and 36.8 grams of oxygen?

2ZnS + 3O2 → 2ZnO + 2SO2

Answer 11

0.208 moles

Question 12

20.30 grams of zinc sulfide (ZnS) is burned in 36.46 grams of oxygen gas to form zinc oxide (ZnO) and sulfur dioxide gas (SO2 ), according to the following balanced chemical equation:
2ZnS + 3O2 → 2ZnO + 2SO2
What is the limiting reagent?

Answer 12

ZnS

Question 13

In the following reaction, what is the limiting reactant if the reaction begins with 7.4 g Ca(OH)2 and 10.4 g H3PO4?

3Ca(OH)2 + 2H3PO4 → Ca3(PO4 )2 + 6H2O

Answer 13

Ca(OH)2