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GCSE & A level Chemistry Calculations: Calculating product volumes of reacting gases Reacting gas volume ratio calculations and Gay-Lussac's Law of combining volumes

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations 10. Reacting gas volume ratios of reactants or products (Avogadro's Law, Gay-Lussac's Law)

Quantitative chemistry calculations Help for problem solving in doing reacting gas volume calculations. Practice revision questions on gas volume ratios of reactants and products from balanced equations, using experiment data, making predictions. This page describes and explains, with fully worked out examples, how to calculate the volumes of gaseous reactants or products formed from given volumes of reactants or products. You need to know how to apply Avogadro's Law and Gay-Lussac's Law of combining volumes. These particular calculation methods simply use the ratio of reactant gases or product gases given by the symbol equation. From reacting gas volumes you can also use the mole concept to calculate the mass of products formed. You can also work out the molecular formula of an unknown gaseous compound from reacting volumes (historical method - much more sophisticated methods these days!). Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses. These revision notes and practice questions on how to do reacting gas volume ratio chemical calculations (using Gay-Lussac's law) and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (9–1) chemistry science courses.

Spotted any careless error? or request a type of GCSE calculation not covered?

Advanced notes on gas law calculations, kinetic model theory of an IDEAL GAS & non-ideal gases

AND Moles and the molar volume of a gas, Avogadro's Law

Self-assessment Quizzes reacting gas volume ratios:  type in answer QUIZ  or  multiple choice QUIZ

Note on two acronyms and associated units:

STP or s.t.p. is a largely defunct abbreviation for 'standard temperature and pressure', 0oC/273K temperature and a pressure of 1 atm/101.3 kPa (101325 Pa). (In bygone times 760 mm of Hg of a 'mercury' barometer/manometer)

NTP or RTP is a frequently used abbreviation for 'normal laboratory temperature and pressure', 25oC/298K temperature and  pressure of 1 atm/101.3 kPa pressure (101325 Pa).

Revision notes on Avogadro's Law, solving reacting gas volume ratio problems, how to calculate gas volumes of products from reactant gas volume ratios, help when revising for AQA A level & GCSE chemistry, Edexcel A level & GCSE chemistry, OCR A level & GCSE gateway science chemistry, OCR GCSE 21st century science chemistry notes for A level & GCSE 9-1 chemistry examination practice questions on Gay-Lussac's Law, working out equations from reacting gas volumes, using reactant product molar ratios to work out gas volumes, relating molar ratios to gas volumes by interpreting the equation 10. Reacting gas volume ratios of reactants or products (Avogadro's Law, Gay-Lussac's Law) In the diagram above, if the volume on the left syringe is twice that of the gas volume in the right, then there are twice as many moles or actual molecules in the left-hand gas syringe.

• REACTING GAS VOLUMES and MOLE RATIO

• Historically Gay-Lussac's Law of volumes states that 'gases combine with each other in simple proportions by volume', but the basis of this reacting gas ratio law, is non other than .

• Avogadro's Law states that 'equal volumes of gases at the same temperature and pressure contain the same number of molecules' or moles of gas.

• Note the mention of the mole concept - in these sorts of calculations you are essentially reading the equation as a mole ratio, so Avogadro's Law supports the prior Gay-Lussac law on the volume ratio of reacting gases and the volume of the products.

• This means the molecule ratio of the equation or the relative moles of reactants and products automatically gives us the gas volumes ratio of reactants and products ...

• ... provided all the gas volumes are measured at the same temperature and pressure.

• These calculations only apply to gaseous reactants or products AND if they are all at the same temperature and pressure.

• The balanced equation can be read/interpreted in terms of either ...

• (i) a gas volume ratio, obviously for gaseous species only (g), AND at the same temperature and pressure.

• or (ii) a mole ratio, which applies to anything in the equation, whether (g), (l) or (s).

• Note: If you have to convert from moles to volume or volume to moles, you need to know the molar volume at that temperature and pressure e.g. 24 dm3 (litres) at 25oC (298K) and 1 atm (101 kPa) pressure.

• i.e. if the volume is in dm3 (litres) at ~ room temperature and pressure

• moles of gas = Vgas/24   or   Vgas = 24 x moles of gas

• If the gas volume is given in cm3, then dm3 = V/1000

• Reacting gas volume ratio calculation Example 10.1

• Given the equation:    HCl(g) + NH3(g) ===> NH4Cl(s)

• 1 mole hydrogen chloride gas combines with 1 mole of ammonia gas to give 1 mole of ammonium chloride solid,

• since from Avogadro's, equal volumes of gases at the same T & P, have the same number of molecules and equal numbers of moles have the same number of molecules, and this gives rise to Gay-Lussac's Law of combining volumes, and we can then logically say directly from the equation ...

• 1 volume of hydrogen chloride will react with 1 volume of ammonia to form solid ammonium chloride

• e.g. 25cm3 + 25cm3 ==> solid product (no gas formed)

• or 400dm3 + 400 dm3 ==> solid product etc.

• so, if 50 cm3 HCl reacts, you can predict 50 cm3 of NH3 will react etc. etc. !

• -

• Reacting gas volume ratio calculation Example 10.2

• Given the equation:    N2(g) + 3H2(g) ==> 2NH3(g)

• 1 mole of nitrogen gas combines with 3 mols of hydrogen gas to form 2 mol of a ammonia gas.

• 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia

• e.g. what volume of hydrogen reacts with 50 cm3 nitrogen and what volume of ammonia will be formed?

• The mole ratio is 1 : 3 ==> 2,

• so you multiply equation ratio numbers by 50 giving ...

• 50 cm3 nitrogen + 150 cm3 hydrogen (3 x 50) ==> 100 cm3 of ammonia (2 x 50)

• -

• Reacting gas volume ratio calculation Example 10.3 (sometimes you need to think 'outside the box' in some context e.g.

• Some industrial processes produce toxic carbon monoxide as a by-product and this gas must be dealt with.

• One way is to burn it to release heat energy for power generation.

• Carbon monoxide burns in air to form carbon dioxide according to the equation:

• 2CO(g)  +  O2(g)  ===>  2CO2(g)

• (a) If carbon monoxide is produced in an industrial process at the rate of 50 dm3 per minute, what rate of oxygen input is required to completely burn it to harmless carbon dioxide?

• The equation reads as 2 mol CO reacts with 1 mol O2 to form 2 mol of CO2

• Therefore 2 volumes of CO reacts with 1 volume of O2 to form 2 volumes of CO2 by ratio.

• Therefore volume of O2 needed = half the volume of CO produced.

• Therefore for every 50 dm3 of CO produced you need 25 dm3 of oxygen to burn it.

• Therefore you need to pump in oxygen at the rate of 25 dm3/min

• (b) However, it is expensive to use oxygen, so air is employed for the combustion process.  If you assume air contains approximately 20% oxygen what rate of air needs to be pumped into the reactor for the carbon monoxide combustion process?

• If air is composed of 20% oxygen, then 1/5th of air is oxygen, so you need five times more air than pure oxygen.

• Therefore rate of air needed = 5 x 25 = 125 dm3 air/min

• -

• Reacting gas volume ratio calculation Example 10.4

• Given the equation:   C3H8(g)  + 5O2(g) ===> 3CO2(g) + 4H2O(l)

• Reading the balanced equation in terms of moles (or mole ratio) ...

• 1 mole of propane gas reacts with 5 mols of oxygen gas to form 3 moles of carbon dioxide gas and 4 mols of liquid water.

• (a) What volume of oxygen is required to burn 25cm3 of propane, C3H8.

• Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5 for burning the fuel propane.

• so actual ratio is 25 : 5x25, so 125cm3 oxygen is needed.

• -

• (b) What volume of carbon dioxide is formed if 5dm3 of propane is burned?

• Theoretical reactant-product volume ratio is C3H8 : CO2 is 1 : 3

• so actual ratio is 5 : 3x5, so 15dm3 carbon dioxide is formed.

• -

• (c) What volume of air (1/5th oxygen) is required to burn propane at the rate of 2dm3 per minute in a gas fire?

• Theoretical reactant volume ratio is C3H8 : O2 is 1 : 5

• so actual ratio is 2 : 5x2, so 10dm3 oxygen per minute is needed,

• therefore, since air is only 1/5th O2,  5 x 10 = 50dm3 of air per minute is required.

• -

• Reacting gas volume ratio calculation Example 10.5

• Given the equation:   2H2(g) + O2(g) ===> 2H2O(l)

• If 40 dm3 of hydrogen, (at 25oC and 1 atm pressure) were burned completely ...

• a) What volume of pure oxygen is required for complete combustion?

• From the balanced equation the reacting mole ratio = gas volume ratio is 2 : 1 for H2 to O2

• Therefore 20 dm3 of pure oxygen is required (40 : 20 is a ratio of 2 : 1).

• -

• b) What volume of air is required if air is ~20% oxygen?

• ~20% is ~1/5, therefore you need five times more air than pure oxygen

• Therefore volume of air needed = 5 x 20 = 100 dm3 of air

• -

• c) What mass of water is formed?

• The easiest way to solve this problem is to think of the water as being formed as a gas-vapour.

• The theoretical gas volume ratio of reactant hydrogen to product water is 1 : 1

• Therefore, prior to condensation at room temperature and pressure, 40 dm3 of water vapour is formed.

• 1 mole of gas occupies 24 dm3, and the relative molar mass of water is 18 g/mol

• (atomic masses H = 1, O = 16, so Mr(H2O) = 1 + 1 + 16 = 18).

• Therefore moles of water formed = 40/24 = 1.666 moles

• Since moles = mass / formula mass

• mass = moles x formula mass

• mass water formed = 1.666 x 18 = 30g of H2O

• -

• Reacting gas volume ratio calculation Example 10.6

• It was found that exactly 10 cm3 of bromine vapour (Br2(g)) combined with exactly 30 cm3 chlorine gas (Cl2(g)) to form a bromine-chlorine compound BrClx.

• a) From the reacting gas volume ratio, what must be the value of x? and hence write the formula of the compound.

• Since both reactants have the same formula, i.e. both diatomic molecules, the ratio of bromine to chlorine atoms in the compound must be 1 : 3 because the reacting gas volume ratio is 1 : 3

• Therefore x must be 3, and the formula must be BrCl3

• -

• b) Write a balanced equation to show the formation of BrClx

• The reacting gas volume ratio is 1 : 3, therefore we can write with certainty that 1 mole (or molecule) of bromine reacts with 3 moles (or molecules) of chlorine, and balancing the symbol equation, results in two moles (or two molecules) of the bromine-chlorine compound being formed in the balanced equation.

• Br2(g) + 3Cl2(g) ==> 2BrCl3(g)

• -

• Reacting gas volume ratio calculation Example 10.7

• If 0.25 moles of ammonia is decomposed,

• (a) how many moles of nitrogen and ammonia will be formed?

• (b) what volume of nitrogen and hydrogen will be formed at 25oC and 1 atm pressure?

• (a) moles of products

• First set out the balanced equation (which may be given in the question)

• 2NH3(g)  ===> N2(g) + 3H2(g)

• From the equation the mole ratio of reactants to products is 2 ==> 1 : 3

• Therefore 0.25 mol ammonia ===>

• 0.125 mol nitrogen (0.25/2) : 0.375 mol hydrogen (3/2 x 0.25)

• -

• (b) volumes of products

• First, convert the moles of ammonia into a gas volume

• Vgas = mol of gas x molar volume

• VNH3 = 0.25 x 24 = 6.0 dm3

• From the mole ratio OR reacting volume ratio (its all the same!) 2 ==> 1 : 3

• 6 dm3 NH3 ===> 3 dm3 N2 (6/2 for nitrogen) : 9 dm3 H2 (3/2 x 6.0 for hydrogen)

• -

• Q's 10.8 and 10.9 are a couple of harder questions, which are really two similar calculations. You would be probably given the equations and the fraction of oxygen in air. Assume all gas volume measurements are made at the same temperature and pressure ..

• Reacting gas volume ratio calculation Example 10.8

• What volume of oxygen/air is required to completely burn a mixture of 10 cm3 of hydrogen and 20 cm3 of carbon monoxide?

• (i) hydrogen: H2(g) + ½O2(g) ===> H2O(l)

• mole ratio/gas volume ratio H2 : O2 is 1 : 0.5

• therefore reacting volume ratio must be the same for 10 cm3 : y cm3

• so y = 10 x 0.5 / 1 = 5 cm3 (10/2)

• (ii) carbon monoxide: CO(g) + ½O2(g) ===> CO2(g)

• mole ratio/gas volume ratio CO : O2 is 1 : 0.5

• therefore reacting volume ratio must be the same for 20 cm3 : z cm3

• so z = 20 x 0.5 / 1 = 10 cm3 (20/2)

• (iii) Therefore total volume oxygen = y + z = 5 + 10 = 15 cm3 O2

• (iv) Assuming air is 1/5th (20%) oxygen, the volume of air required would be 5 x 15 = 75 cm3 air

• -

• Reacting gas volume ratio calculation Example 10.9

• (this is a definitely a bit more tricky on the equations, but check out 10.7 first)

• What volume of oxygen/air is required to completely burn a mixture of 20 dm3 of methane and 10 dm3 of propane?

• (i) methane: CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l)

• mole ratio = gas volume ratio CH4 : O2 is 1 : 2

• therefore reacting volume ratio is 20 : y

• so y = 20 x 2 / 1 = 40 dm3

• (ii) propane: C3H8(g)  + 5O2(g) ==> 3CO2(g) + 4H2O(l)

• mole ratio = gas volume ratio C3H8 : O2 is 1 : 5

• therefore reacting volume ratio is 10 : z

• so z = 10 x 5 / 1 = 50 dm3

• (iii) Therefore total volume oxygen = y + z = 40 + 50 = 90 dm3 O2

• (iv) Assuming air is 1/5th (20%) oxygen, the volume of air required would be 5 x 90 = 450 dm3 air

• -

• Reacting gas volume ratio calculation Example 10.10

• This is a limiting reactant question, but again assume all volumes are measured at the same temperature and pressure.

• Fluorine and chlorine are both very reactive gases, but fluorine, from the group 7 halogen reactivity trend, is the most reactive of the two gases. Fluorine reacts with chlorine to form chlorine(III) fluoride.

• The balanced equation is:  3F2(g)  +  Cl2(g)  ===>  2ClF3(g)

• If 200 cm3 of fluorine gas is reacted with 50 cm3 of chlorine gas:

• (a) What is the theoretical gas volume ratio for the reactants?

• The molar ratio given by the balanced equation above is 3 : 1,

• so the theoretical reacting gas volume ratio is 3 : 1 (for fluorine : chlorine)

• (b) Which is the limiting reactant?

• The molar ratio = the theoretical reacting gas volume ratio of 3 : 1,

• but the actual volume ratio of the initial mixture is 200 : 50 or 4 : 1 (fluorine : chlorine)

• Since 4/1 exceeds 3/1 the excess reactant is fluorine and therefore chlorine is the limiting reactant.

• (c) What volume of chlorine(III) fluoride is formed?

• The molar ratio of chlorine to product is 1 : 2 (Cl2 : ClF3)

• Therefore the reacting gas volume ratio is also 1 : 2

• Therefore 50 cm3 of chlorine will be converted to 50 x 2/1 = 100 cm3 of ClF3.

• (d) Calculate the volume of unreacted fluorine.

• mole ratio = volume ratio = 3 : 2 for F2 : ClF3 (selected reactant : product)

• Since 100 cm3 of ClF3 was formed, from a ratio of 3 : 2 (3/2),

• the volume of fluorine consumed must be 3/2 x 100 = 150 cm3.

• Therefore the unreacted fluorine volume = the initial 200 - 150 = 50 cm3 of excess unreacted fluorine.

• Note: You can also work it out from just the reactant gas volume ratio of 3 : 1 (F2 : Cl2 3/1)

• The limiting reactant volume of 50 cm3 of chlorine must therefore react with 50 x 3/1 = 150 cm3 F2

• thus leaving 200 - 150 = 50 cm3 of unreacted fluorine gas.

• -

• 10.11

Self-assessment Quizzes on REACTING GAS VOLUME RATIO CALCULATIONS Gay-Lussac's Law

or

Advanced notes on gas law calculations, kinetic model theory of an IDEAL GAS & non-ideal gases

Moles and the molar volume of a gas, Avogadro's Law Above is typical periodic table used in GCSE science-chemistry specifications in doing reacting gas volume ratio calculations, and I've 'usually' used these values in my exemplar calculations to cover most syllabuses

Revision notes on Avogadro's Law, solving reacting gas volume ratio problems, how to calculate gas volumes of products from reactant gas volume ratios, help when revising for AQA A level & GCSE chemistry, Edexcel A level & GCSE chemistry, OCR A level & GCSE gateway science chemistry, OCR GCSE 21st century science chemistry notes for A level & GCSE 9-1 chemistry examination practice questions on Gay-Lussac's Law, working out equations from reacting gas volumes, using reactant product molar ratios to work out gas volumes, relating molar ratios to gas volumes by interpreting the equation

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