CHAPTER 9

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CHAPTER 9

Stoichiometry

More Conversions!

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Stoichiometry

• Needs a balanced equation

• Use the balanced eqn to predict ending and / or starting amounts

• Coefficients are now mole ratios

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Mole Ratios

Example: 2Al₂O₃(l) → 4Al(s) + 3O₂(g) Mole Ratios: 2 mol Al₂O₃

4 mol Al

2 mol Al₂O₃ 3 mol O₂

4 mol Al 3 mol O₂

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Mole Ratios

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Mole – Mole Problem A

CO₂(g) + 2LiOH(s) → Li₂CO₃(s) + H₂O(l) How many moles of lithium hydroxide are

required to react with 20 mol CO₂, the average amount exhaled by a person each day?

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Mole – Mole Problem A

20 mol X mol

CO₂(g) + 2LiOH(s) → Li₂CO₃(s) + H₂O(l)

How many moles of lithium hydroxide are required to react with 20 mol CO₂, the average amount exhaled by a person each day?

Step 1:

Start with what you know from the problem.

20 mol CO₂ = X mol LiOH

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Mole – Mole Problem A

20 mol X mol

1 mol 2 mol

Step 2:

Determine what you know from the balanced equation.

1 mol CO₂ = 2 mol LiOH

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Mole – Mole Problem A

Step 3:

Set-up so units cancel & solve.

20 mol CO₂ = X mol LiOH 1 mol CO₂ 2 mol LiOH

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Mole – Mole Problem A

X = 40 mol LiOH

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Mole – Mole Problem B

5.3 mol X mol

CO₂(g) + 2LiOH(s) → Li₂CO₃(s) + H₂O(l) How many moles of lithium carbonate are

produced when 5.3 mol CO₂are reacted?

Step 1:

5.3 mol CO₂ = X mol Li₂CO₃

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Mole – Mole Problem B

5.3 mol X mol

1 mol 1 mol

How many moles of lithium carbonate are produced when 5.3 mol CO₂are reacted?

Step 2:

Determine what you know from the balanced eqn.

1 mol CO₂ = 1 mol Li₂CO₃

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Mole – Mole Problem B

5.3 mol X mol

1 mol 1 mol

Step 3:

Set-up units to cancel out & solve.

5.3 mol CO₂ = X mol Li₂CO₃ 1 mol CO₂ 1 mol Li₂CO₃

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Mole – Mole Problem B

X = 5.3 mol Li₂CO₃

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Mole – Mass Problems

mol – g

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mol – Mass Problems

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mol – Mass Problems

1. Start with what you know.

2. Write a ratio of what you know from the problem above the equation.

3. Write a ratio of what you know from the balanced equation below.

4. Set the ratios together and Solve.

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Balanced Equation:

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

What is the mass of glucose (C₆H₁₂O₆) produced from 3.00 mol of water (H₂O)?

mol – Mass Problem C

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Balanced Equation:

3.00 mol X grams

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

mol – Mass Problem C

Step 1:

3.00 mol H₂O = X g C₆H₁₂O₆

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Balanced Equation:

3.00 mol X grams

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

6 mol 1mol*(180.18g/mol)

mol – Mass Problem C

Step 2:

Use the balanced equation to find the mol of known and the grams of unknown.

6 mol H₂O = 180.18 g C₆H₁₂O₆

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Balanced Equation:

3.00 mol X grams

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

6 mol 180.18 g

mol – Mass Problem C

Step 3:

Set-up and solve.

3.00 mol H₂O = X g C₆H₁₂O₆

6 mol H₂O 180.18 g C₆H₁₂O₆

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Balanced Equation:

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

mol – Mass Problem C

6 X = (3.00)(180.18)

6 6

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Balanced Equation:

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

mol – Mass Problem C

X = (3.00)(180.18) 6

X = 90.09 g C₆H₁₂O₆

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Balanced Equation:

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

What is the mass of oxygen (O₂) produced from 2.50 mol of water (H₂O)?

mol – Mass Problem D

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Balanced Equation:

2.50 mol X grams

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

mol – Mass Problem D

Step 1:

2.50 mol H₂O = X g O₂

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Balanced Equation:

2.50 mol X grams

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

^{6 mol} (6mol * 32g/mol = 192g)

mol – Mass Problem D

Step 2:

6 mol H₂O = 192.00 g O₂

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Balanced Equation:

2.50 mol X grams

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

^{6 mol} (6 * 32 = 192g)

mol – Mass Problem D

Step 3:

Set-up and solve.

2.50 mol H₂O = X g O₂

6 mol H₂O 192.00 g O₂

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Balanced Equation:

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

mol – Mass Problem D

6 X = (2.50)(192.00)

6 6

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Balanced Equation:

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)

mol – Mass Problem D

X = 80.0 g O₂

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Mass – mol Problems

g - mol

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Mass – mol Problems

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Mass – mol Problems

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Balanced Equation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

How many moles of NO are formed from 824 g of NH₃?

Mass – mol Problem E

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Balanced Equation:

^{824 g} ^{X mol}

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Mass – mol Problem E

Step 1:

824 g NH₃= X mol

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Balanced Equation:

824 g X mol

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

(4mol*17.01g/mol = 68.04g) 4 mol

Mass – mol Problem E

Step 2:

68.04 g NH₃ = 4 mol NO

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Balanced Equation:

824 g X mol

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

(4*17.01 = 68.04) 4 mol

Mass – mol Problem E

Step 3:

Set-up your ratio.

824 g NH₃ = X mol NO 68.04 g NH₃ 4 mol NO

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Balanced Equation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Mass – mol Problem E

824 g NH₃ = X mol NO 68.04 g NH₃ 4 mol NO

68.04 X = (4)(824) 68.04 68.04

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Balanced Equation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Mass – mol Problem E

X = 48.4 g NO

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Balanced Equation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

How many moles of H₂O are formed from 412 g of NH₃?

Mass – mol Problem F

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Balanced Equation:

412 g X mol

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Mass – mol Problem F

Step 1:

412 g NH₃ = X mol H₂O

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Balanced Equation:

412 g X mol

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

(4*17.03 = 68.12) 6 mol

Mass – mol Problem F

Step 2:

Determine what is known from the balanced equation.

68.12 g NH₃ = 6 mol H₂O

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Balanced Equation:

412 g X mol

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

(4*17.03 = 68.12) 6 mol

Mass – mol Problem F

Step 3:

Set the ratios together and Solve.

412 g NH₃ = X mol H₂O 68.12 g NH₃ 6 mol H₂O

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Balanced Equation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Mass – mol Problem F

68.12 X = (412)(6) 68.12 68.12

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Balanced Equation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Mass – mol Problem F

X = 36.3 g H₂O

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Mass – Mass Problems

g - g

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Gram – Gram Problems

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Gram – Gram Problems

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Gram – Gram Problems

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Gram – Gram Problem G

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

How many grams of SnF₂ are produced from the reaction of 30.00 g HF?

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Gram – Gram Problem G

30.00 g X g

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

Step 1:

30.00 g HF = X g SnF₂

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Gram – Gram Problem G

30.00 g X g

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

(2mol*20.01g/mol=40.0g) 1mol*156.69g/mol

Step 2:

40.02 g HF = 156.69 g SnF₂

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Gram – Gram Problem G

30.00 g X g

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

(2*20.01=40.02) 156.69

How many grams of SnF₂ are

produced from the reaction of 30.00 g HF?

Step 3:

Put the ratios together and Solve.

30.00 g HF = X g SnF₂ 40.02 g HF 156.69 g SnF₂

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Gram – Gram Problem G

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

40.02 X = (30.00)(156.69) 40.02 40.02

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Gram – Gram Problem G

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

X = 117.5 g SnF₂

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Gram – Gram Problem H

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g) How many grams of HF are produced from the reaction of 150.5 g H₂?

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Gram – Gram Problem H

X g 150.5 g

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

How many grams of HF are produced from the reaction of 150.5 g H₂?

Step 1:

150.5 g H₂= X g HF

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Gram – Gram Problem H

X g 150.5 g

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

(2*20.01=40.02) (2.02)

Step 2:

2.02 g H₂ = 40.02 g HF

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Gram – Gram Problem H

X g 150.5 g

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

(2*20.01=40.02) (2.02)

Step 3:

Set-up the ratios and Solve.

150.5 g H₂ = X g HF 2.02 g H₂ 40.02 g HF

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Gram – Gram Problem H

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

2.02 X = (150.5)(40.02) 2.02 2.02

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Gram – Gram Problem H

Sn(s) + 2HF(g) → SnF₂(s) + H₂(g)

X = 2982 g HF

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To Review …

When solving:

Known Amount & Unit Given in Problem

Known Amount &

Unit from Equation

=

Unknown Unit Given in Problem

Unknown Amount &

Unit from Equation