IB Colourful Solutions in Chemistry

Standard level

Gases, involved in chemical reactions show simple ratios between gas volumes. This makes dealing with gaseous reactions very simple in terms of calculations.

Syllabus ref: S1.4.6

Structure 1.4.6 - Avogadro’s law states that equal volumes of all gases measured under the same conditions of temperature and pressure contain equal numbers of molecules.

Solve problems involving the mole ratio of reactants and/or products and the volume of gases.

Guidance

Tools and links

Structure 1.5 - Avogadro’s law applies to ideal gases.
Under what conditions might the behaviour of a real gas deviate most from an ideal gas?

Gay Lussac's law
Avogadro's hypothesis
Worked examples
Now test yourself

Gay Lussac's law

In 1808 the Freanch chemist Gay Lussac investigating the reactions of gases came to the conclusion that when gases combine chemically they do so in volumes that have a simple ratio to one another, and to any gaseous product, provided that all gases are measured at the same temperature and pressure. This became known as Gay Lussac's law.

For example, in the reaction between hydrogen and oxygen making water:

2 volumes of hydrogen

1 volume of oxygen

hydrogen + oxygen → water
2H₂ + O₂ → 2H₂O

The volume of hydrogen needed for complete reaction is always double the volume of oxygen, provided the temperature and pressure of the two gases are the same.

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Avogadro's hypothesis

Extending Gay Lussac's work in 1811, Avogadro suggested that "equal volumes of all gases contain equal number of molecules (the gases being measured at the same temperature and pressure). This became known as Avogadro's hypothesis or Avogadro's law.

This means that in the reaction:

N₂ + 3H₂ → 2NH₃

For a given volume of nitrogen, three times the volume of hydrogen is needed for complete reaction. The volume of nitrogen contains a certain number of molecules and there are three times as many molecules in the volume of hydrogen.

N₂	+	3H₂	→	2NH₃
1 molecule	+	3 molecules	→	2 molecules
1 volume	+	3 volumes	→	2 volumes

Avogadro's hypothesis allows us to substitute the coefficients of any balanced gaseous equation for volumes of gas, for example:

Example: Find the volume of hydrogen required to react completely with 200 cm³ of nitrogen according to the equation:

N₂ + 3H₂ → 2NH₃

The equation tells us that 1 volume of nitrogen reactrs completely with 3 volumes of hydrogen:

Therefore volume of hydrogen = 3 x volume of nitrogen

Therefore volume of hydrogen = 3 x 200 cm³ = 600 cm³

Example:

Using the equation:

N₂ + 3H₂ → 2NH₃

if 100 cm³ of nitrogen was used (this is now equivalent to 1 'volume') it would need 3 times as much hydrogen for complete reaction i.e. 3 volumes = 300 cm³.

The reaction would produce 2 volumes of ammonia i.e. 2 x 100 cm³ = 200 cm³

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Worked examples

Q146-01 In the reaction between sulfur dioxide and oxygen forming sulfur(VI) oxide, the gases are passed over a hot vanadium pentoxide catalyst acording to the equation:

2SO₂ + O₂ → 2SO₃

Calculate the volume of oxygen needed to react fully with 200 cm³ of sulfur dioxide.
Answer

The equation shows that sulfur dioxide reacts with oxygen in a 2:1 ratio

Therefore 2 volumes of sulfur dioxide react with 1 volume of oxygen

Thus for complete reaction 200 cm³ of sulfur dioxide needs 100 cm³ of oxygen

Q146-02 What volume of carbon dioxide will be formed from the complete combustion of 200 cm³ methane?

Answer

Methane burns according to the equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

1 volume of methane produces 1 volume of carbon dioxide

Therefore 200cm³ methane produces 200cm³ of carbon dioxide

Q146-03 What volume of oxygen will be required to completely oxidise 200 cm³ of ammonia according to the equation:

4NH₃ + 5O₂ → 4NO + 6H₂O

Answer

From the equation 4 volumes of ammonia requires 5 volumes of oxygen

Therefore 200 cm³ of ammonia requires 200 x 5/4 volumes of oxygen = 250 cm³ oxygen

Q146-04 What will be the final volume of gas when 300 cm³ ethene reacts with 300 cm³ oxygen at 400ºC?

Answer

Equation for the reaction:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Inspection of the equation shows that 1 volume of ethene requires 3 volumes of oxygen. In the question there is insufficient oxygen to react with all the ethene, i.e. the oxygen is the limiting reagent and only 100 cm³ of ethene can react.

Therefore from the equation 1 volume of ethene reacts with 3 volumes of oxygen to produce 2 volumes of carbon dioxide and 2 volumes of steam (at this temperature)

So 100 cm³ of ethene reacts with 3 x 100 cm³ of oxygen to produce 200 cm³ of carbon dioxide and 200cm³ of steam

Initial volume of gases = 300 cm³ + 300 cm³

Final volume of gases = 200 cm³(CO2) + 200 cm³(H2O) + 200 cm³ (unreacted methane) = 600 cm³ of gas mixture

Q146-05 What will be the final volume of the gas mixture, when 100cm³ propane reacts with 600 cm³ oxygen at 400ºC?

Answer

Equation for the reaction:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Inspection of the equation shows that 1 volume of propane requires 5 volumes of oxygen. In the question there is insufficient propane to react with all of the oxygen, i.e. the propane is the limiting reagent and only 500 cm³ of oxygen can react.

Therefore from the equation 1 volume of propane reacts with 5 volumes of oxygen to produce 3 volumes of carbon dioxide and 4 volumes of steam (at this temperature)

So 100 cm³ of propane reacts with 5 x 100cm³ of oxygen to produce 300cm³ of carbon dioxide and 400 cm³ of steam

Initial volume of gases = 100 cm³ + 600 cm³ = 700 cm³

Final volume of gases = 300cm³(CO2) + 400cm³ (H2O) + 100cm³ (unreacted oxygen) = 800 cm³ of gas mixture

Q146-06 When 100 cm³ of ethene is burned in excess oxygen, calculate the volume of carbon dioxide produced (all gas volumes measured at STP)

Answer

Equation for the reaction:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

1 volume of ethene makes 2 volumes of CO₂

100cm³ of ethene = 100/22700 moles = 4.46 x 10^-3 moles

this makes 2 x 4.46 x 10^-3 moles of carbon dioxide = 8.93 x 10^-3 moles

This is equivalent to 8.93 x 10^-3 x 22700 = 200cm³

Note: the final answer demonstrates Gay Lussac's law of combining gas volumes - moles are proportional to gas volumes

Q146-07 Calculate the volume of oxygen required to react with excess carbon monoxide, to produce 500 cm³ of carbon dioxide gas.

Answer

The equation for the reaction:

2CO + O₂ → 2CO₂

Thus, 2 volumes of carbon monoxide react with 1 volume of oxygen to produce 2 volumes of carbon dioxide.

If 500 cm³ of carbon dioxide gas are required then 2 volumes = 500 cm³

1 volume of oxygen is required = 1 x 500/2 = 250cm³ of oxygen

Q146-08 In the combusion of methanal, 100 cm³ of methanal reacted with 200 cm³ of oxygen. Calculate the composition of the gas mixture produced, if all measurements were taken at STP.
Answer

Equation for the reaction:

HCHO + O₂ → CO₂ + H₂O

Thus, 1 volume of methanal reacts with 1 volume of oxygen in a complete reaction.

100 cm³ of methanal will react with only 100 cm³ of of oxygen leaving 100 cm³ of oxygen unused.

The final composition of the gas mixture = 100 cm³ of oxygen + 100 cm³ of carbon dioxide (water is a liquid at STP)

Q146-09 At 200ºC carbon disulfide vapour reacts with oxygen according to the equation:

CS₂ + 3O₂ → CO₂ + 2SO₂

Calculate the final composition of the gas mixture produced when 1.2 dm³ of carbon disulfide vapour reacts with 1.2 dm³ of oxygen gas

Answer

From the equation:

1 volume of carbon disulfide reacts completely with 3 volumes of oxygen

In this case the oxygen is the limiting reagent and only 1.2/3 dm³ = 400 cm³ of the carbon disulfide can react (leaving 800 cm³ of the carbon disulfide unreacted)

400 cm³ of carbon disulfide (equivalent to 1 volume in the equation) produces 400 cm³ of carbon dioxide and 2 x 400 cm³ = 800 cm³ of sulfur dioxide.

The final mixture = 400 cm³ of carbon dioxide, 800cm³ of sulfur dioxide and 800 cm³ of unreacted carbon disulfide

Q146-10 In a gunpowder mixture, the ratio of sulfur to carbon is 3 :1 by mass. If both are fully oxidised in the explosion to sulfur and carbon dioxide and 200cm³ of sulfur dioxide are formed, calculate the volume of carbon dioxide that is also produced in the explosion.

Answer

sulfur and carbon react with oxygen according to the equations:

S + O₂ → SO₂

C + O₂ → CO₂

1 mole of gas is produced per mole of both carbon and sulfur

Relative atomic mass of S = 32 and C = 12

moles = mass/relative mass

Therefore mole ratio of sulfur to carbon = 3/32 : 1/12

multiply through by 12 giving sulfur to carbon ratio = 36/32 : 1

Thus if 200cm³ sulfur dioxide are formed then 200 x 32 / 36 cm³ = 177.8 cm³ of carbon dioxide are formed

Q146-11 Calculate the moles of gas present at STP in a 5 dm³ flask.

Answer

1 moles of gas occupies a volume of 22700 cm³ at STP

Therefore 5 dm³ = 5000cm³ contains 5000/22700 moles = 0.223 moles at STP

Q146-12 What volume of oxygen corresponds to 0.2 moles, measured at STP.

Answer

1 mole occupies a volume = 22700 cm³

Therefore 0.2 moles = 0.2 x 22700 cm³ = 4480cm³

Q146-13 Find the volume occupied at STP, by a mixture containing 0.05 moles of hydrogen and 0.025 moles of oxygen gas.

Answer

Total moles of gas = 0.05 + 0.025 = 0.075 moles

1 mole occupies a volume = 22700 cm³

Therefore 0.075 moles of gas occupy a volume of 0.075 x 22700 cm³ = 1680 cm³

Q146-14 Calculate the number of moles of gas contained at STP in a 100 cm³ gas syringe.

Answer

1 mole occupies a volume = 22700 cm³ at STP

Therefore 100 cm³ is equivalent to = 100/22700 moles of gas = 4.46 x 10^-3

Q146-15 Find the volume occupies at STP by a mixture containing 0.01 moles of neon and 0.03 moles of oxygen gas.

Answer

Total moles of gas = 0.01 + 0.03 = 0.04

1 mole of gas occupies 22700 cm³ at STP

Therefore 0.04 moles occupies 0.04 x 22700 cm³ = 896cm³

Q146-16 Find the amount of oxygen corresponding to a volume at STP of 40 cm³ of oxygen gas.

Answer

Oxygen, O₂

At STP 40cm³ = 40/22700 moles = 1.79 x 10^-3 moles

Therefore 40cm³ oxygen corresponds to = 1.79 x 10^-3 moles

Q146-17 A mixture of 400cm³ of oxygen gas and 640 cm³ of hydrogen gas is contained at STP, calculate the total number of moles present.
Answer

Total volume of gas = 400cm³ of oxygen gas + 640 cm³ of hydrogen gas = 1040 cm³

At STP 1 mole occupies 22700 cm³

Therefore number of moles = 1040/22700 = 0.0458 moles

Q146-18 If 2 moles of carbon monoxide gas are mixed with and 4 moles of hydrogen gas at STP, calculate the volume occupied.
Answer

Total moles of gas = 2 + 4 = 6 moles

1 mols of gas occupies 22700 cm³ at STP

Therefore volume occupied = 6 x 22.7 dm³ = 136.2 dm³

Q146-19 A sample of natural gas contains 120 moles of methane and 20 moles of helium. Calculate the volume in m³ occupied by the gases at STP to three significant figures.
Answer

Total moles of gas = 120 + 20 = 140 moles

1 mole occupies 22.7 dm³ at STP, therefore 140 moles occupies 22.7 x 140 = 3178 dm³

1000dm³ = 1m³

Volume occupied by the natural gas mixture = 3178/1000 = 3.18 m³

Q146-20 A sample of syngas contains 60 moles of carbon monoxide, 15 moles of carbon dioxide and 60 moles of hydrogen. Calculate the volume in m³ occupied by the gases at STP, to three significant figures.
Answer

Total moles of gas = 60 + 15 + 15 = 90 moles

1 mole occupies 22.7 dm³ at STP, therefore 90 moles occupies 22.7 x 90 = 2043 dm³

1000dm³ = 1m³

Therefore the volume occupied by the syngas mixture = 2043/1000 = 2.04 m³

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