Title: Lesson 8: Calculations involving Volumes of Gases

Learning Objectives:•Understand that a fixed quantity (in moles) of gas, always occupies the same volume at room temperature•Perform calculations using the molar volume of a perfect gas

Refresh•Chloroethene, C2H3Cl, reacts with oxygen according to the equation below.

2C2H3Cl(g) + 5O2(g) → 4CO2(g) + 2H2O(g) + 2HCl(g)

What is the amount, in mol, of H2O produced when 10.0 mol of C2H3Cl and 10.0 mol of O2 are mixed together, and the above reaction goes to completion?

A. 4.00B. 8.00C. 10.0D. 20.0

•Equal masses of the metals Na, Mg, Ca and Ag are added to separate samples of excess HCl(aq). Which metal produces the greatest total volume of H2(g)?

A. NaB. MgC. CaD. Ag

• The number of particles in the two flasks are the same.

• Even though bromine molecules are larger and heavier, this is not relevant because of the nature of gaseous state. We assume that the volume of individual gas molecule are zero.

• Particles in a gas are widely spaced out with negligible forces between them. Most gas space is EMPTY. We call this an IDEAL GAS.

• Therefore gas volume is determined only by the number of particles and the temperature and pressure.

How do the number of particles in the two flasks compare?

Avogadro’s lawIn 1811 the Italian scientist Amedeo Avogadro developed a theory about the volume of gases.

Avogadro’s law:Equal volumes of different gases at the same pressure and

temperature will contain equal numbers of particles.

For example, if there are 2 moles of O2 in 50 cm3 of oxygen gas, then there will be 2 moles of N2 in 50 cm3 of nitrogen gas and 2 moles of CO2 in 50 cm3 of carbon dioxide gas at the same temperature and pressure.

Using this principle, the volume that a gas occupies will depend on the number of moles of the gas.

Alternatively, it can be stated that equal numbers of particles of all gases, when measured at the same temperature and pressure, occupy equal volumes.

• The volume occupied by a gas is known as the MOLAR VOLUME

• This can be used in a similar way to using MOLAR MASS but these calculations are easier because gases have the same molar volume under the same conditions

The Molar Volume of an Ideal Gas

• At standard temperature and pressure (STP):– Molar Volume of Gas = 22.7 dm3 mol-1

= 2.27x10-2 m3 mol-1

– T = 273K (0oC)

– P = 1.01x105 Pa (100kPa)

Molar volumes of gasesIf the temperature and pressure are fixed at convenient standard values, the molar volume of a gas can be determined.

Standard temperature is 273 K and pressure is 100 kPa.

At standard temperature and pressure, 1 mole of any gas occupies a volume of 22.7 dm3. This is the molar volume.

Example: what volume does 5 moles of CO2 occupy?

volume occupied = no. moles × molar volume

= 5 × 22.7

= 113.5 dm3

In Calculations….

• What volume of H2 gas is produced when 0.050 mol Li reacts with excess Hydrochloric acid at S.T.P.?

• Write balanced equation– 2 Li(s) + 2 HCl(aq) 2 LiCl(aq) + H2(g)

– Work out the mole ratio:– 2 moles Li to 1 mole H2

– 0.050 mol Li reacts to form 0.025 mol of H2

• Plug into the formula• V = 0.025 x 22.7 = 0.567 dm3

More calculations … with limiting reagents• At STP, 30 cm3 ethane reacts with 60 cm3 oxygen. Which reactant is present

in excess, how much remains after the reaction and what volume of CO2 is produced?– 2 C2H6(g) + 10 O2(g) 6 H2O(l) + 4 CO2(g)

– Limiting reactant:• C2H6: 30/2 = 15

• O2: 60/10 = 6– therefore O2 limiting, ‘6’ will be the number used in all further calculations as there is enough O2 for ‘6’ of

the reaction

– C2H6 remaining:• V(C2H6 used) = 6 x 2 = 12 cm3

• V(C2H6 remaining) = 30 – 12 = 18 cm3 EXCESS AFTER REACTION

– Note: there is no need to convert to moles as they are all in a ratio of moles as you would divide by 22.7 to get to moles, do your sums and then multiply by 22.7 to get back to volumes… So why bother?

Volume CO2 produced: V(CO2) = 6 x 4 = 24 cm3

Time to practice• What is the minimum volume of H2 gas required to fully reduce 10.0 g copper (II) oxide to

copper (assume STP)

CuO(s) + H2(g) Cu(s) + H2O(l)

• In a car airbag, sodium azide (NaN3) decomposes explosively to make N2 gas. What is the minimum mass of sodium azide required to fully inflate a 60.0 dm3 airbag, assuming STP?

2 NaN3(s) 2 Na(s) + 3 N2(g)

• 500 cm3 carbon monoxide reacts with 300 cm3 oxygen to produce carbon dioxide. What are the final volumes of each of the three gases on completion of the reaction?

CO(g) + O2(g) CO2(g)

Key Points

• The volume of gas depends on the temperature, pressure and number of moles, NOT THE TYPE OF GAS

• Molar Volume at STP = 22.7 dm3