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study examples carefully10. Reacting gas volume ratios of reactants or products (Avogadro's Law, Gay-Lussac's Law)

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!).

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study examples carefully10. Reacting gas volume ratios of reactants or products (Avogadro's Law, Gay-Lussac's Law)

gas syringes

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.

  • Historically Gay-Lussac's Law of volumes states that 'gases combine with each other in simple proportions by volume'.

  • 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.

  • 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, if 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.

  • 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.

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

    • e.g. 25cm3 + 25cm3 ==> solid product

    • or 400dm3 + 400 dm3 ==> solid product (no gas formed)

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

  • Example 10.3: Given the equation:   C3H8(g)  + 5O2(g) ==> 3CO2(g) + 4H2O(l)

    • 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

  • Example 10.4: 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 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

  • Example 10.5: It was found that exactly 10 cm3 of bromine vapour (Br2(g)) combined with exactly 30 cm3 chlorine gas (Cl2(g)) to form 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 molecules) of the bromine-chlorine compound being formed.

        • Br2(g) + Cl2(g) ==> 2BrCl3(g)

  • Example 10.6:

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OTHER CALCULATION PAGES

  1. What is relative atomic mass?, relative isotopic mass and calculating relative atomic mass

  2. Calculating relative formula/molecular mass of a compound or element molecule

  3. Law of Conservation of Mass and simple reacting mass calculations

  4. Composition by percentage mass of elements in a compound

  5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

  6. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination

  7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

  8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

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

  10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (this page)

  11. Molarity, volumes and solution concentrations (and diagrams of apparatus)

  12. How to do volumetric titration calculations e.g. acid-alkali titrations (and diagrams of apparatus)

  13. Electrolysis products calculations (negative cathode and positive anode products)

  14. Other calculations e.g. % purity, % percentage & theoretical yield, volumetric titration apparatus, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

  15. Energy transfers in physical/chemical changes, exothermic/endothermic reactions

  16. Gas calculations involving PVT relationships, Boyle's and Charles Laws

  17. Radioactivity & half-life calculations including dating materials

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