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, C 2 H 3 Cl, reacts with oxygen according to the equation below. 2C 2 H 3 Cl(g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O(g) + 2HCl(g) What is the amount, in mol, of H 2 O produced when 10.0 mol of C 2 H 3 Cl and 10.0 mol of O 2 are mixed together, and the above reaction goes to completion? A.4.00 B.8.00 C.10.0 D.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 H 2 (g)? A.Na B.Mg C.Ca D.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 law In 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 O 2 in 50 cm 3 of oxygen gas, then there will be 2 moles of N 2 in 50 cm 3 of nitrogen gas and 2 moles of CO 2 in 50 cm 3 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 dm 3 mol -1 = 2.27x10 -2 m 3 mol -1 – T = 273K (0 o C) – P = 1.01x10 5 Pa (100kPa)

Molar volumes of gases If 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 dm 3. This is the molar volume. Example: what volume does 5 moles of CO 2 occupy? volume occupied = no. moles × molar volume = 5 × 22.7 = dm 3

In Calculations…. What volume of H 2 gas is produced when mol Li reacts with excess Hydrochloric acid at S.T.P.? Write balanced equation – 2 Li(s) + 2 HCl(aq)  2 LiCl(aq) + H 2 (g) – Work out the mole ratio: – 2 moles Li to 1 mole H 2 – mol Li reacts to form mol of H 2 Plug into the formula V = x 22.7 = dm 3

More calculations … with limiting reagents At STP, 30 cm 3 ethane reacts with 60 cm 3 oxygen. Which reactant is present in excess, how much remains after the reaction and what volume of CO 2 is produced? – 2 C 2 H 6 (g) + 10 O 2 (g)  6 H 2 O(l) + 4 CO 2 (g) – Limiting reactant: C 2 H 6 : 30/2 = 15 O 2 : 60/10 = 6 – therefore O 2 limiting, ‘6’ will be the number used in all further calculations as there is enough O 2 for ‘6’ of the reaction – C 2 H 6 remaining: V(C 2 H 6 used) = 6 x 2 = 12 cm 3 V(C 2 H 6 remaining) = 30 – 12 = 18 cm 3 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 CO 2 produced:  V(CO 2 ) = 6 x 4 = 24 cm 3

Time to practice What is the minimum volume of H 2 gas required to fully reduce 10.0 g copper (II) oxide to copper (assume STP) CuO(s) + H 2 (g)  Cu(s) + H 2 O(l) In a car airbag, sodium azide (NaN 3 ) decomposes explosively to make N 2 gas. What is the minimum mass of sodium azide required to fully inflate a 60.0 dm 3 airbag, assuming STP? 2 NaN 3 (s)  2 Na(s) + 3 N 2 (g) 500 cm 3 carbon monoxide reacts with 300 cm 3 oxygen to produce carbon dioxide. What are the final volumes of each of the three gases on completion of the reaction? CO(g) + O 2 (g)  CO 2 (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 dm 3