- Given the following symbol equation:
**2Mg + O**(Relative atomic masses, A_{2}==> 2MgO

Calculate how many g of magnesium oxide is formed by burning 24g of magnesium in air._{r}:**Mg = 24 and O = 16**) [com-1]- 40
- 16
- 20
- 80

- Given the following symbol equation:
**2Mg + O**(Relative atomic masses, A_{2}==> 2MgO

Calculate how many tonne of magnesium is needed to make 80 tonne of magnesium oxide._{r}:**Mg = 24 and O = 16**) [com-2]- 96
- 48
- 24
- 16

- Given the following symbol equation:
**CaCO**(Relative atomic masses, A_{3(s)}==> CaO_{(s)}+ CO_{2(g)}

Calculate how many kg of calcium carbonate is needed to make 56 kg of calcium oxide._{r}:**Ca = 40, C = 12 and O = 16**) [com-3]- 50
- 44
- 100
- 40

- Given the following symbol equation:
**CaCO**(Relative atomic masses, A_{3(s)}==> CaO_{(s)}+ CO_{2(g)}

Calculate how many g of carbon dioxide is formed if 25g of calcium carbonate is decomposed on heating to form 14g of calcium oxide._{r}:**Ca = 40, C = 12 and O = 16**) [com-4]- 22
- 44
- 10
- 11

- Given the symbol equation to show the formation of iron sulphide:
**Fe + S ==> FeS**(Relative atomic masses, A

Calculate the mass in g of iron sulphide formed when 5.6g of iron combines with 3.2g of sulphur._{r}:**Fe = 56 and S = 32**) [com-5]- 8.8
- 2.4
- 5.6
- 3.2

- Given the symbol equation to show the formation of iron sulphide:
**Fe + S ==> FeS**(Relative atomic masses, A

Calculate the mass in g of iron sulphide formed when 28g of iron combines with 16g of sulphur._{r}:**Fe = 56 and S = 32**) [com-6]- 12
- 44
- 28
- 16

- Given the symbol equation to show the formation of calcium chloride by burning calcium in chlorine:
**Ca**(Relative atomic masses, A_{(s)}+ Cl_{2(g)}==> CaCl_{2(s)}

Calculate the mass in g of calcium chloride formed when 20g of calcium combines with 35.5g of chlorine._{r}:**Ca = 40 and Cl = 35.5**) [com-7]- 15.5
- 70.0
- 55.5
- 17.5

- Given the symbol equation to show the formation of calcium chloride by burning calcium in chlorine:
**Ca**(Relative atomic masses, A_{(s)}+ Cl_{2(g)}==> CaCl_{2(s)}

Calculate the mass in g of chlorine needed when 40g of calcium forms 111g of calcium chloride._{r}:**Ca = 40 and Cl = 35.5**) [com-8]- 80
- 100
- 151
- 71

- Given the symbol equation to show the formation of aluminium sulphide by heating a mixture of aluminium and sulphur:
**2Al + 3S ==> Al**(Relative atomic masses, A_{2}S_{3}

How many g of sulphur is needed if 54g of aluminium is reacted to form 150g of aluminium sulphide?_{r}:**Al = 27 and S = 32**) [com-9]- 96
- 204
- 64
- 27

- Given the symbol equation to show the formation of aluminium sulphide by heating a mixture of aluminium and sulphur:
**2Al + 3S ==> Al**(Relative atomic masses, A_{2}S_{3}

How many kg of sulphur is needed if 108kg of aluminium is reacted to form 300kg of aluminium sulphide?_{r}:**Al = 27 and S = 32**) [com-10]- 408
- 192
- 96
- 32

- Given the symbol equation to show the formation of iron sulphide by heating a mixture of iron and sulphur:
**Fe + S ==> FeS**(Relative atomic masses, A

Calculate the mass in g of iron unreacted when 60g of iron reacts with 32g of sulphur to form 88g of iron sulphide._{r}:**Fe = 56 and S = 32**) [com-11]- 8
- 32
- 4
- 56

- Given the symbol equation to show the formation of iron sulphide by heating a mixture of iron and sulphur:
**Fe + S ==> FeS**(Relative atomic masses, A

Calculate the mass in g of sulphur unreacted when 28g of iron reacts with 22g of sulphur to form 44g of iron sulphide._{r}:**Fe = 56 and S = 32**) [com-12]- 22
- 16
- 12
- 6

- Given the symbol equation to show the formation of calcium chloride by burning calcium in chlorine:
**Ca**(Relative atomic masses, A_{(s)}+ Cl_{2(g)}==> CaCl_{2(s)}

Calculate the mass in g of calcium left unreacted when 25g of calcium reacts with 35.5g of chlorine._{r}:**Ca = 40 and Cl = 35.5**) [com-13]- 10.5
- 5.0
- 7.1
- 12.5

- Given the symbol equation to show the formation of calcium chloride by burning calcium in chlorine:
**Ca**(Relative atomic masses, A_{(s)}+ Cl_{2(g)}==> CaCl_{2(s)}

Calculate the mass in g of chlorine left unreacted when 80g of calcium reacts with 150g of chlorine to form 222g of calcium chloride._{r}:**Ca = 40 and Cl = 35.5**) [com-14]- 70
- 8
- 16
- 72