Stoichiometry Practice Sheet

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Stoichiometry Practice Sheet. Init 1/10/2012 by Daniel R. Barnes.

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Stoichiometry Practice Sheet

Init 1/10/2012 by Daniel R. Barnes

WARNING: This presentation may contain graphical images and other content that was taken from the world wide web without permission of the owner(s). Please do not copy or distribute this presentation. Its very existence may be illegal.

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

1. Draw a cartoon of the reaction below, represeting each atom as a circle with its element’s symbol inside.

O

O

C

O

C

H

H

H

O

C

O

O

C

C

C

H

H

O

O

H

H

H

H

H

O

O

O

H

H

H

O

O

H

O

O

O

O

O

O

H

H

O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

1

2. How many oxygen molecules are consumed for every propane molecule burned?

used up

A: five

Would looking at the cartoon help?

H

H

H

O

O

O

C

C

C

C

C

C

H

H

O

O

O

O

H

H

H

O

O

O

H

H

H

H

O

O

O

O

O

O

H

H

O

H

H

O

O

O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

1

2. How many oxygen molecules are consumed for every propane molecule burned?

A: five

Would looking at the cartoon help?

One molecule of propane

H

H

H

O

O

O

C

C

C

C

C

C

H

H

O

O

O

O

H

H

H

4

O

O

O

1

3

H

H

H

H

O

O

O

O

O

O

5

2

H

H

O

H

H

O

O

O

Five molecules of oxygen

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

1

3. How many carbon dioxide molecules are produced for every propane molecule burned?

A: three

Would looking at the cartoon help?

H

H

H

O

O

O

C

C

C

C

C

C

H

H

O

O

O

O

H

H

H

O

O

O

H

H

H

H

O

O

O

O

O

O

H

H

O

H

H

O

O

O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

1

3. How many carbon dioxide molecules are produced for every propane molecule burned?

Three molecules of carbon dioxide

A: three

One molecule of propane

H

H

H

O

O

O

C

C

C

C

C

C

H

H

O

O

O

O

1

2

3

H

H

H

O

O

O

H

H

H

H

O

O

O

O

O

O

H

H

O

H

H

O

O

O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

4. How many water molecules are created for every oxygen molecule consumed?

4 water molecules

=

0.8 water molecules/oxygen molecule

5 oxygen molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

5. If two million propane molecules are burned, how many water molecules are produced?

4

water molecules

2 million propane molecules

x

propane molecule

1

=

8 million water molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

6. How many oxygen molecules does it take to produce nine billion carbon dioxide molecules?

5

oxygen molecules

9 billion carbon dioxide molecules

x

3

carbon dioxide molecules

1

=

15 billion oxygen molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

7. How many propane molecules does it take to produce 484 dozen water molecules?

1

propane molecules

484 dozen water molecules

x

4

water molecules

1

=

121 dozen propane molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

8. How much carbon dioxide would be produced by burning

10 kazillion oxygen molecules?

3

carbon dioxide molecules

10 kazillion oxygen molecules

x

5

oxygen molecules

1

=

6 kazillion carbon dioxide molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

9. For every mole of carbon dioxide produced, how many moles of water are produced?

4

mol water

1 mol carbon dioxide

x

3

mol carbon dioxide

1

=

1.33 mol water

You could also say, “1.33 mol water/mol carbon dioxide.”

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

10. What is the oxygen-to-carbon dioxide ratio in this reaction?

5

oxygen molecules

5 to 3

=

5:3

=

3

carbon dioxide molecules

5

mol oxygen

=

3

mol carbon dioxide

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:

C3H8 + 5O2 3CO2 + 4H2O

11. In general, how do you figure out the mole ratio of any two substances in a chemical reaction?

The coefficients

COEFFICIENTS MOLE RATIO

And now, for the

BACK SIDE

of the worksheet

(Sorry. The rest of the front side is under construction for now.)

12. What mass of magnesium oxide should be produced when 12.15 grams of magnesium metal burns? Magnesium burns according to the following equation: 2Mg + O2 2MgO

g

20

12.15 g

2Mg + O2 2MgO

12.15 g Mg

1

mol Mg

2

mol MgO

40

g MgO

1

24

g Mg

2

mol Mg

1

mol MgO

Mg:

MgO:

Mg:

1

x 24 =

24

Mg:

1

x 24 =

24

O:

1

x 16 =

16

24

g/mol

40

g/mol

960

12.15 x 2 x 40

g MgO

g MgO

g MgO

=

20

=

48

24 x 2

12. What mass of magnesium oxide should be produced when 12.15 grams of magnesium metal burns? Magnesium burns according to the following equation: 2Mg + O2 2MgO

g

20

12.15 g

2Mg + O2 2MgO

12.15 g Mg

1

mol Mg

2

mol MgO

40

g MgO

1

24

g Mg

2

mol Mg

1

mol MgO

Mg:

MgO:

Mg:

1

x 24 =

24

Mg:

1

x 24 =

24

O:

1

x 16 =

16

24

g/mol

40

g/mol

960

12.15 x 2 x 40

g MgO

g MgO

g MgO

=

20

=

48

24 x 2

13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H2 + O2 2H2O.

g

16

128 g

2H2 + O2 2H2O

128 g O2

1

mol O2

2

mol H2

2

g H2

1

32

g O2

1

mol O2

1

mol H2

O2:

H2:

O:

2

x 16 =

32

H:

2

x 1 =

2

2

g/mol

32

g/mol

512

128 x 2 x 2

g H2

g H2

g H2

=

16

=

32

32

13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H2 + O2 2H2O.

g

16

128 g

2H2 + O2 2H2O

128 g O2

1

mol O2

2

mol H2

2

g H2

1

32

g O2

1

mol O2

1

mol H2

O2:

H2:

O:

2

x 16 =

32

H:

2

x 1 =

2

2

g/mol

32

g/mol

512

128 x 2 x 2

g H2

g H2

g H2

=

16

=

32

32

14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO3 CuO + CO2 ?

g

400

620 g

CuCO3  CuO + CO2

620 g CuCO3

mol CuCO3

1

mol CuO

80

g CuO

1

1

124

g CuCO3

1

mol CuCO3

1

mol CuO

CuO:

CuCO3:

Cu:

1

x 64 =

64

Cu:

1

x 64 =

64

C:

1

x 12 =

12

O:

1

x 16 =

16

O:

3

x 16 =

48

80

g/mol

124

g/mol

49,600

620 X 80

g CuO

g CuO

g CuO

=

400

=

124

124

14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO3 CuO + CO2 ?

g

400

620 g

CuCO3  CuO + CO2

620 g CuCO3

1

mol CuCO3

1

mol CuO

80

g CuO

1

124

g CuCO3

1

mol CuCO3

1

mol CuO

CuO:

CuCO3:

Cu:

1

x 64 =

64

Cu:

1

x 64 =

64

C:

1

x 12 =

12

O:

1

x 16 =

16

O:

3

x 16 =

48

80

g/mol

124

g/mol

49,600

620 X 80

g CuO

g CuO

g CuO

=

400

=

124

124

15. How many grams of NI3 must explode in order to produce 7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI3 N2 + 3I2

g

7900

7620 g

2NI3 N2 + 3I2

7620 g I2

mol I2

2

mol NI3

395

g NI3

1

1

254

g I2

3

mol I2

1

mol NI3

NI3:

I2:

I:

2

x 127 =

254

N:

1

x 14 =

14

I:

3

x 127 =

381

254

g/mol

395

g/mol

7620 x 2 x 395

6,019,800

g NI3

g NI3

g NI3

=

7900

=

762

254 x 3

15. How many grams of NI3 must explode in order to produce 7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI3 N2 + 3I2

g

7900

7620 g

2NI3 N2 + 3I2

7620 g I2

mol I2

2

mol NI3

395

g NI3

1

1

254

g I2

3

mol I2

1

mol NI3

NI3:

I2:

I:

2

x 127 =

254

N:

1

x 14 =

14

I:

3

x 127 =

381

254

g/mol

395

g/mol

7620 x 2 x 395

6,019,800

g NI3

g NI3

g NI3

=

7900

=

762

254 x 3

HONORS PROBLEM

(From the Q2 BMK, given 1/28-29/2014, by Action Learning Systems)

• Iron reacts with oxygen to form rust,
• 4Fe (s) + 3O2 (g)  2Fe2O3 (s).
• When 56 g of iron (Fe) reacts with 96 g of oxygen (O2), how much rust (Fe2O3) will form?

This is a limiting reactant stoichiometry problem. It’s stoichiometry because you are given grams of one chemical and asked to determine grams of some other chemical. It’s a limiting reactant problem because you are actually given grams for TWO chemicals, iron and oxygen.

The way to solve this is to do two stoichiometry calculations. First, calculate how much rust can be made from 56 grams of iron. Then, calculate how much rust can be made from 96 grams of oxygen. The answer is the lesser of the two results, since oxygen and iron must cooperate to make rust, and a chain is only as strong as its weakest link.

96 g O2

56 g Fe

mol O2

mol Fe

2

2

mol Fe2O3

mol Fe2O3

160

160

g Fe2O3

g Fe2O3

1

1

1

1

56

32

g O2

g Fe

3

4

mol O2

mol Fe

1

1

mol Fe2O3

mol Fe2O3

HONORS PROBLEM

(From the Q2 BMK, given 1/28-29/2014, by Action Learning Systems)

• Iron reacts with oxygen to form rust,
• 4Fe (s) + 3O2 (g)  2Fe2O3 (s).
• When 56 g of iron (Fe) reacts with 96 g of oxygen (O2), how much rust (Fe2O3) will form?

Limiting reactant

= 80 g Fe2O3

Excess reactant

= 320 g Fe2O3

If you think there’s something wrong or missing,

barnesd@centinela.k12.ca.us

TM

HYPERINDEX

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Q2 BMK #38 bonus question

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