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Gases for AQA A-level Chemistry

Gases

This page covers the following topics:

1. Ideal gas equation
2. Volume of gases
3. Volume of reactants and products
4. Mass changes

The ideal gas equation relates pressure, volume, temperature and amount of gas. The equation is pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. It is important to use the SI units of all the variables in the equation: pressure is given in pascals (Pa), volume is given in mยณ, the gas constant is 8.31 J Kโปยน molโปยน, and the temperature needs to be given in kelvin (K).

Ideal gas equation

According to Avogadro's Law, one mole of any gas occupies the same volume (at the same temperature and pressure). At a temperature of 25ยฐC and a pressure of 101.3 kPa (these conditions are also referred to as rtp), that volume is 24 dmยณ. 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.

Volume of gases

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 dmยณ of carbon monoxide reacts with oxygen to produce carbon dioxide. The balanced equation is 2CO + Oโ‚‚ โ†’ 2COโ‚‚, which gives us the molar ratio 2 : 1 โ†’ 2. Therefore, the volume of carbon dioxide produced will be equal to the volume of carbon monoxide (25dmยณ), and the volume of oxygen will be half the volume of carbon monoxide, i.e. 12.5 dmยณ.

Volume of reactants and products

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.

Mass changes

1

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

Number of moles = 5/2 = 2.5. V = 2.5 ร— 24 = 60 dmยณ.

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

2

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

Number of moles = 7/44 = 0.16. V = 0.16 ร— 24 = 3.84 dmยณ.

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

3

How many moles of molecules does a sample of gas with a volume of 0.57 mยณ at 101000 Pa and 300 K contain?

pV = nRT
n = pV/RT
n = (101000 ร— 0.57)/(8.31 ร— 300) = 23 mol

23 mol

How many moles of molecules does a sample of gas with a volume of 0.57 mยณ at 101000 Pa and 300 K contain?

4

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

Hโ‚‚ + Clโ‚‚ โ†’ 2HCl. Molar ratio is 1 : 1 โ†’ 2. Therefore, the volume of Hโ‚‚ = the volume of Clโ‚‚ = half the volume of HCl = 19.8 dmยณ.

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

5

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?

2Feโ‚‚Oโ‚ƒ + 3Cโ†’ 4Fe + 3COโ‚‚. There are 57/160 = 0.36 moles of Feโ‚‚Oโ‚ƒ present. The molar ratio is 2 : 3 โ†’ 4 : 3. Therefore, 0.54 moles of COโ‚‚ will escape into the air. m = n ร— Mแตฃ = 0.54 ร— 44 = 23.76 g. Thus, the total mass will decrease by 23.76 g.

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?

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