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Notes Chapter 9 Limiting Reagent Sample Problems Page 1

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Problem 1: Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas. Suppose that 6.70 mol Na reacts with 3.20 mol Cl2.

A. What is the limiting reagent?

B. How many moles of NaCl are produced?

2 Na + Cl2 → 2 NaCl

To answer question A,

1. Write the mole amounts under their respective places.

2 Na + Cl2 → 2 NaCl

6.7 mol 3.2 mol

2. Pick one of these and work the stoichiometry toward the other one. (You can either work from the Na to the Cl2, or work from the Cl2 to the Na. The choice is yours.)

2 Na + Cl2 → 2 NaCl (I choose to work from Na to Cl2)

6.7 mol 3.2 mol

6.7 mol Na 1 mol Cl2 = 3.35 mol of Cl2 needed for this reaction to take place.

2 mol Na

3. Now determine if the problem gave you enough Cl2 for this reaction to take place.

No they didn’t. We need 3.35 moles of Cl2, and they only gave us 3.2 moles of Cl2. Since we don’t have enough Cl2, it must be our limiting reagent.

4. So, the limiting reagent for this reaction is the Cl2.

To answer question B,

1. Place an “LR” over the limiting reagent and work the stoichiometry from the limiting reagent over to what the problem wants to know (How many moles of NaCl are produced?)

LR

2 Na + Cl2 → 2 NaCl

6.7 mol 3.2 mol ? mol

3.2 mol Cl2 2 mol NaCl = 6.40 mol NaCl are produced 1 mol Cl2

(2)

Problem 2: The equation for the complete combustion for ethane (C2H4) is written below. If 2.70 mol C2H4 is reacted with 6.30 mol O2,

A. Identify the limiting reagent.

B. Calculate the moles of water produced.

C2H4 + 3 O2 → 2 CO2 + 2 H2O

To answer question A,

1. Write the mole amounts under their respective places.

C2H4 + 3 O2 → 2 CO2 + 2 H2O

2.70 mol 6.30 mol

2. Pick one of these and work the stoichiometry toward the other one. (You can either work from the C2H4 to the O2, or work from the O2 to the C2H4. The choice is yours.)

C2H4 + 3 O2 → 2 CO2 + 2 H2O (I choose to work from C2H4 to O2)

2.70 mol 6.30 mol

2.70 mol C2H4 3 mol O2 = 8.1 mol of O2 needed for this reaction to take place.

1 mol C2H4

3. Now determine if the problem gave you enough O2 for this reaction to take place.

No they didn’t. We need 8.1 moles of O2, and they only gave us 6.3 moles of O2. Since we don’t have enough O2, it must be our limiting reagent.

4. So, the limiting reagent for this reaction is the O2.

To answer question B,

1. Place an “LR” over the limiting reagent and work the stoichiometry from the limiting reagent over to what the problem wants to know (How many moles of H2O are produced?)

LR

C2H4 + 3 O2 → 2 CO2 + 2 H2O

2.70 mol 6.30 mol ? mol

6.3 mol O2 2 mol H2O = 4.2 mol H2O are produced 3 mol O2

(3)

Problem 3: The equation for the incomplete combustion for ethane (C2H4) is written below.

If 2.70 mol C2H4 is reacted with 6.30 mol O2,

C. Identify the limiting reagent.

D. Calculate the moles of water produced.

C2H4 + 2 O2 → 2 CO + 2 H2O (notice we have different products.)

To answer question A,

1. Write the mole amounts under their respective places.

C2H4 + 2 O2 → 2 CO + 2 H2O

2.70 mol 6.30 mol

2. Pick one of these and work the stoichiometry toward the other one. (You can either work from the C2H4 to the O2, or work from the O2 to the C2H4. The choice is yours.)

C2H4 + 2 O2 → 2 CO + 2 H2O (I choose to work from C2H4 to O2)

2.70 mol 6.30 mol

2.70 mol C2H4 2 mol O2 = 5.4 mol of O2 needed for this reaction to take place.

1 mol C2H4

3. Now determine if the problem gave you enough O2 for this reaction to take place.

Yes they did. We only need 5.4 moles of O2, and they gave us 6.3 moles of O2. Since the problem gave us more than enough O2, the “other guy (the C2H4) must be our limiting reagent.

4. So, the limiting reagent for this reaction is the C2H4.

To answer question B,

1. Place an “LR” over the limiting reagent and work the stoichiometry from the limiting reagent over to what the problem wants to know (How many moles of H2O are produced?)

LR

C2H4 + 3 O2 → 2 CO2 + 2 H2O

2.70 mol ? mol

2.7 mol C2H4 2 mol H2O = 5.4 mol H2O are produced 1 mol C2H4

(4)

Problem 4: A. What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

B. What is the maximum number of grams of Cu2S that can be formed?

2 Cu + S → Cu2S

To answer question A,

1. Write the mole amounts under their respective places.

2 Cu + S → Cu2S

80.0 g 25.0 g

2. Pick one of these and work the stoichiometry toward the other one. (You can either work from the Cu to the S, or work from the S to the Cu. The choice is yours.)

2 Cu + S → Cu2S (I choose to work from Cu to S)

80.0 g 25.0 g

mol mol

80.0 g Cu 1 mol Cu 1 mol S 32 g S = 20.0 g S needed for this 64 g Cu 2 mol Cu 1 mol S reaction to take place.

3. Now determine if the problem gave you enough S for this reaction to take place.

Yes they did. We only need 20.0 g of S, and they gave us 25.0 g of S. Since the problem gave us more than enough S, the “other guy (the Cu) must be our limiting reagent.

4. So, the limiting reagent for this reaction is the Cu.

To answer question B,

1. Place an “LR” over the limiting reagent and work the stoichiometry from the limiting reagent over to what the problem wants to know (How many grams of Cu2S are produced?)

LR

2 Cu + S → Cu2S (Work from the LR over to the Cu2S.)

80.0 g ? g

mol mol

80.0 g Cu 1 mol Cu 1 mol Cu2S 160 g Cu2S = 100 g of Cu2S produced 64 g Cu 2 mol Cu 1 mol Cu2S

(5)

Problem 5: Hydrogen gas can be produced in the laboratory by the reaction of magnesium metal with hydrochloric acid.

A. Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg.

B. How many grams of hydrogen can be produced when 6.00 g HCl is added to 5.00 g Mg?

Mg + 2 HCl → MgCl2 + H2

To answer question A,

1. Write the mole amounts under their respective places.

Mg + 2 HCl → MgCl2 + H2

5.0 g 6.0 g

2. Pick one of these and work the stoichiometry toward the other one. (You can either work from the Mg to the HCl, or work from the HCl to the Mg. The choice is yours.)

Mg + 2 HCl → MgCl2 + H2 ( I choose to work from Mg to HCl)

5.0 g 6.0 g

mol mol

5.0 g Mg 1 mol Mg 2 mol HCl 36.5 g HCl = 15.2 g HCl needed for 24 g Mg 1 mol Mg 1 mol HCl this reaction to take place.

3. Now determine if the problem gave you enough HCl for this reaction to take place.

No they didn’t. We need 15.2 g of HCl, and they only gave us 6.0 g of HCl.

Since we don’t have enough HCl, it must be our limiting reagent.

4. So, the limiting reagent for this reaction is the HCl.

To answer question B,

1. Place an “LR” over the limiting reagent and work the stoichiometry from the limiting reagent over to what the problem wants to know (How many grams of H2 are produced?)

LR

Mg + 2 HCl → MgCl2 + H2 ( I choose to work from Mg to HCl)

6.0 g ? g

mol mol

6.0 g HCl 1 mol HCl 1 mol H2 2 g H2 = 0.16 g H2 are produced.

36.5 g HCl 2 mol HCl 1 mol HCl

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