 # Gases for OCR GCSE Chemistry 1. Volume of gases
2. Volume of reactants and products
3. Mass changes

According to Avogadro's Law, one mole of any gas occupies the same volume (at the same temperature and pressure). At a temperature of 25°C and a pressure of 101.3 kPa (these conditions are also referred to as rtp), that volume is 24 dm³. The volume of a gas can thus be calculated by first calculating its number of moles (the mass of the gas over its relative formula mass) and then multiplying the number of moles by the volume of 1 mole. As gases with the same number of moles have the same volume, the volume of gaseous reactants and products can be calculated using the volume of one reactant or product and the balanced equation of the reaction. Example: 25 dm³ of carbon monoxide reacts with oxygen to produce carbon dioxide. The balanced equation is 2CO + O₂ → 2CO₂, which gives us the molar ratio 2 : 1 → 2. Therefore, the volume of carbon dioxide produced will be equal to the volume of carbon monoxide (25dm³), and the volume of oxygen will be half the volume of carbon monoxide, i.e. 12.5 dm³. According to the law of conservation of mass, the total mass of the products in a reaction will equal the total mass of the reactants. However, if the reaction takes place in a non-enclosed system and involves a gas as a reactant or product, mass changes may seem to occur. If one of the products is a gas, the mass may decrease as some of the gas may escape into the air. If one of the reactants is a gas that occurs in the air, the total mass increases because the gas was not present in the reaction flask at the start of the reaction. # 1

Calculate the volume in dm³ of a sample of hydrogen weighing 5 g at rtp.

Number of moles = 5/2 = 2.5. V = 2.5 × 24 = 60 dm³. # 2

A sample of carbon dioxide weighs 7 g at rtp. Calculate its volume in dm³.

Number of moles = 7/44 = 0.16. V = 0.16 × 24 = 3.84 dm³. # 3

What volumes of hydrogen and chlorine are necessary to produce 39.6 dm³ of hydrogen chloride?

H₂ + Cl₂ → 2HCl. Molar ratio is 1 : 1 → 2. Therefore, the volume of H₂ = the volume of Cl₂ = half the volume of HCl = 19.8 dm³. # 4

What is the mass change that can be observed when reacting 57 g of iron oxide with carbon to produce iron and carbon dioxide in a non-enclosed reaction?

2Fe₂O₃ + 3C→ 4Fe + 3CO₂. There are 57/160 = 0.36 moles of Fe₂O₃ present. The molar ratio is 2 : 3 → 4 : 3. Therefore, 0.54 moles of CO₂ will escape into the air. m = n × Mᵣ = 0.54 × 44 = 23.76 g. Thus, the total mass will decrease by 23.76 g. # 5

Sulfur dioxide is produced by the reaction of sulfur with oxygen from the air. Find the mass of sulfur dioxide produced by reacting 86.4 g of sulfur.

S + O₂ → SO₂. There are 86.4/32 = 2.7 moles of sulfur present. Molar ratio is 1 : 1 → 1, so 2.7 moles of oxygen from the air are required for the reaction. m = n × Mᵣ = 2.7 × 32 = 86.4. Therefore, the mass of SO₂ = 86.4 + 86.4 = 172.8 g. End of page