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CGB IGCSE Chemistry
Practice Problem Answers - the mole 2
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Practice Problem Answers

The Mole 2

 

  1. When carbon burns in oxygen, carbon dioxide is formed.

C(s) + O2(g) CO2(g)

    1. Calculate the mass of 1 mole of carbon.

 

Relative atomic mass of carbon = 12 Ar

1 mole of carbon = 12 g

 

    1. Calculate the mass of 1 mole of oxygen.

 

Relative atomic mass of oxygen = 16 Ar

Relative formula mass of oxygen = 16 Ar x 2 = 32 Mr

1 mole of O2 = 32 g

 

    1. Calculate the mass of 1 mole of carbon dioxide.

 

Relative atomic mass of carbon = 12 Ar

Relative atomic mass of oxygen = 16 Ar

Relative formula mass of carbon dioxide =

C = 12 x 1 = 12

2 O = 16 x 2 = 32

12 + 32 = 44 Mr

1 mole of CO2 = 44 g

 

    1. If 3.4 moles of carbon is burned in excess oxygen, how many moles of carbon dioxide are formed.

 

1 mole C forms 1 mole CO2

therefore

3.4 moles C will form 3.4 moles CO2

 

    1. Find out how much carbon dioxide is formed if 24 g of carbon is burned in excess oxygen.

 

1 mole C = 12 g

24 g C x 1 mole / 12 g C = 2 moles

24 g C = 2 moles C

1 mole C forms 1 mole CO2

therefore

2 moles C forms 2 moles CO2

1 mole CO2 = 44 g

2 moles CO2 x 44g / mole = 88 g CO2

88g of CO2 are formed when 24 g of carbon is burned.

 

    1. Find out how many grams of carbon need to be burned to form 35.2 grams of carbon dioxide.

 

1 mole CO2 = 44 g

35.2 g x 1 mole / 44 g = 0.8 moles

35.2 g CO2 = 0.8 moles CO2

1 mole C forms 1 mole CO2

therefore

0.8 moles C forms 0.8 moles CO2

1 mole C = 12 g

0.8 moles x 12 g / mole = 9.6 g

0.8 moles C = 9.6 g C

you need to burn 9.6 g of C to form 35.2 grams of CO2

 

    1. How many moles of carbon and oxygen need to react to form 13 moles of carbon dioxide.

 

1 mole C + 1 mole O2 1 mole CO2

13 mole C + 13 mole O2 13 mole CO2

You need 13 moles of carbon and 13 moles of oxygen to form 13 moles of CO2.

 

  1. Hydrogen burns in oxygen to form water.

2H2(g) + O2(g) 2H2O(l)

    1. Calculate the mass of 1 mole of hydrogen.

 

Relative atomic mass of hydrogen = 1 Ar

Relative formula mass of hydrogen = 1 Ar x 2 = 2 Mr

1 mole H2 = 2 g

 

    1. Calculate the mass of 1 mole of water.

 

Relative formula mass of H2O

2H = 2 x 1 Ar = 2

O = 1 x 16 Ar = 16

2 + 16 = 18 Mr

                        1 mole H2O = 18 g

 

    1. How many moles of hydrogen and oxygen need to be burned to form 6 moles of water?

 

2 moles H2 + 1 mole O2 2 moles H2O

 

2 moles H2O x 3 = 6 moles H2O

2 moles H2 x 3 = 6 moles H2

1 mole O2 x 3 = 3 moles O2

 

You need to burn 6 moles of H2 and 3 moles of O2 to form 6 moles of water.

    1. Find out how much water is formed if 5 moles of hydrogen is burned in excess oxygen.

 

2 moles H2 + 1 mole O2 2 moles H2O

 

2 moles H2 x 2.5 = 5 moles H2

2 moles H2O x 2.5 = 5 moles H2O

 

5 moles of H2O are formed if 5 moles of hydrogen are burned in excess oxygen.

 

    1. Find out how many grams of water is formed if 28 g of hydrogen is burned in excess oxygen.

 

1 mole H2 = 2 g

28 g x 1 mole / 2g = 14 moles H2

 

2 moles H2 + 1 mole O2 2 moles H2O

 

2 moles H2 x 7 = 14 moles H2

2 moles H2O x 7 = 14 moles H2O

 

1 mole H2O = 18 g

14 moles H2O x 18 g / 1 mole = 252 g H20

 

252 g of H20 are formed if 28 g of hydrogen are burned in excess oxygen.

 

  1. When aqueous solutions of sodium chloride and silver nitrate are mixed, a white precipitate of silver chloride forms.

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

    1. Calculate the mass of 1 mole of silver nitrate.

 

Relative formula mass of AgNO3

Ag = 108 Ar

N = 14 Ar

3 O = 3 x 16 Ar = 48

108 + 14 + 48 = 170 Mr

1 mole of AgNO3 = 170 g

 

    1. Calculate the mass of 1 mole of sodium chloride.

 

Relative formula mass of NaCl

Na = 23 Ar

Cl = 35.5 Ar

23 + 35.5 = 58.5 Mr

1 mole of NaCl = 58.5 g

 

    1. Calculate the mass of 1 mole of silver chloride.

 

Relative formula mass of AgCl

Ag = 108 Ar

Cl = 35.5 Ar

108 + 35.5 = 143.5 Mr

1 mole of AgCl = 143.5 g

 

    1. Calculate the mass of 1 mole of sodium nitrate.

 

Relative formula mass of NaNO3

Na = 23 Ar

N = 14 Ar

O = 16 Ar x 3 = 48

23 + 14 + 48 = 85 Mr

1 mole NaNO3 = 85 g

 

    1. How many moles of silver nitrate and sodium chloride have to react to form 7.5 moles of silver chloride?

 

1 mole AgNO3(aq) + 1 mole NaCl(aq) 1 mole AgCl(s) + 1 mole NaNO3(aq)

 

1 mole AgCl x 7.5 = 7.5 moles AgCl

1 mole AgNO3 x 7.5 = 7.5 moles AgNO3

1 mole NaCl x 7.5 = 7.5 moles NaCl

 

7.5 moles AgNO3 and 7.5 moles NaCl react to form 7.5 moles of AgCl

 

    1. If 11.7 grams of sodium chloride are added to excess silver nitrate how many grams of silver chloride are formed?

 

1 mole NaCl = 58.5

11.7 g x 1 mole / 58.5 g = 0.2 moles NaCl

 

             1 mole AgNO3(aq) + 1 mole NaCl(aq) 1 mole AgCl(s) + 1 mole NaNO3(aq)

 

            1 mole NaCl x 0.2 = 0.2 moles NaCl

            1 mole AgCl x 0.2 = 0.2 moles AgCl

           

            1 mole AgCl = 143.5 g

            0.2 moles x 143.5 g / 1 mole = 28.7 g AgCl

 

28.7 g of AgCl are formed when 11.7 g of sodium chloride are added to excess silver nitrate.

 

    1. If 255 grams of silver nitrate are added to excess sodium chloride how many grams of silver chloride are formed?

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