1. Moles and reactions At the end of this section, you should be able to deduce the quantities of reactants and products from balanced equations

2. Key information Stoichiometry is the molar relationship between the relative quantities of substances taking part in a reaction. Balanced equation gives information about: * the reacting quantities that are needed to prepare a required quantity of a product * the quantity of products formed by reacting together of known quantities of reactants. In dealing with reacting quantities, the following steps should be considered 1. Write the balanced reaction equation if not given 2. Work out the moles of the known reacting quantities 3. Work out the mass or any other quantity required

3. Recall There are three ways to work out an amount in moles 1. From mass: n = m or m = n x M M 2. From gas volumes: n = V(dm 3 ) or V(dm 3 ) = n x 24.0 24.0 3. From solutions: n = c x V(dm 3 )

4. Examples (reacting masses) 1. 8.00g of aluminium reacted with excess oxygen to produce aluminium oxide. a) write a balanced equation b) find the number of moles of aluminium oxide formed. c) what was the mass of aluminium oxide produced? Answer a) 4Al + 3O 2  2Al 2 O 3 4mol 3 mol 2 mol 2 mol 1.5 mol 1 mol b) Moles of aluminium, n = m M

5. n = 8.00 = 0.2963 27 From the reaction, 2 mols of Al produces 1 mol of Al 2 O 3 Therefore 0.2963 mols of Al will produce = 0.2963. 2 = 0.1482mols Al 2 O 3. c) mass = n x M M(Al 2 O 3 ) = 27 x 2 + 16 x3 = 102 mass of Al 2 O 3 produced = 102 x 0.1482 = 15.12g

6. 2. What mass of copper oxide is obtained by heating 4.94g of copper carbonate? CuCO 3 (s)  CuO(s) + CO 2 (g) ( hint. Find the moles, mole ratio of quantities and the mass) Answer from n = m ÷ M M(CuCO 3 ) =64 + 12 + 16 x 3 = 124 n = 4.94 ÷ 124 = 0.0398 mol of CuCO 3 From the reaction, 1 mol of CuCO 3 produces 1 mol of CuO Therefore 0.0398 mol of CuCO 3 produces 0.0398 mol of CuO mass of CuO = n x M M(CuO) = 64 + 16 = 80 = 0.0398 x 80 = 3.18g of CuO

7. 3. Sodium and chorine react to form sodium chloride: 2Na (s)+ Cl 2 (g)  2NaCl(s) (a) Calculate the amount, in mol, in 2.925 g NaCl. n = m/M = 2.925 = 2.925 = 0.0500 mol 23.0 + 35.5 58.5 (b) Calculate the amounts, in mol, of Na and Cl 2 that would form 2.925 g NaCl. equation: 2Na (s)+ Cl 2 (g)  2NaCl(s) amounts in equation 2 mol : 1 mol : 2 mol Amounts required : 0.050mol 0.025mol 0.050mol (c) Calculate the masses of Na and Cl2 that would form 2.925 g NaCl. Na : mass = n x M = 0.050 x 23 = 1.15 g Cl 2 : mass = n x M = 0.025 x (35.5+35.5) = 0.025 + 71 = 1.775 g

8. (4) Zinc carbonate decomposes to produces carbon dioxide and 4.2 g zinc oxide. (a) write a balanced reaction equation (b) find the moles of zinc carbonate that decomposed (c) what mass of zinc carbonate produced the 4.20 g of zinc oxide.

9. Answers (a) Equation: ZnCO 3  ZnO + CO 2 (b) n = m M M(ZnO) = 65.4 + 16.0 = 81.4 mole of ZnO, n = 4.20 = 0.0516 mol 81.4 reaction equation: ZnCO 3  ZnO + CO 2 amounts in equation: 1 mol: 1 mol: 1 mol amounts required:0.0516 0.0516 0.0516 therefore amount of zinc carbonate that decomposed was 0.0516 mol (c)mass = n x M M(ZnCO 3 ) = 65.4 + 12.0 + (16.0x3) = 125.4 mass of zinc carbonate = 125.4 x 0.0516 = 6.47 g therefore 6.47 g zinc carbonate is required

10. Reacting masses and gas volumes If a substance is a solid, mass is used to measure the amount needed for a reaction. However, if a reactant or product is a gas, it unlikely it could be weighed; it would be easier to measure its volume. According to Avogadro’s Law: Equal volumes of gases measured at the same temperature and pressure contain the same number of molecules. This is extremely useful when relating the volume of gas to an equation.

11. Example What volume of oxygen reacts exactly with 100cm 3 of hydrogen at the same temperature and pressure? Answer the equation for the reaction is: 2H 2 (g) + O 2 (g)  2H 2 O(l) the mole ratio 2 mol: 1 mol: 2 mol volume required 100cm 3 50cm 3 100cm 3 The equation indicates that half the number of oxygen molecules as hydrogen molecules is needed. Therefore, 50cm 3 of oxygen reacts with 100cm 3 of hydrogen

12. 2. What volume of oxygen reacts exactly with 30 cm 3 of methane and what volume of carbon dioxide is obtained at the same temperature and pressure? Answer The equation for the reaction is: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) the mole ratio is 1 mol: 2 mol: 1 mol: 2 mol volume of gas 30cm 3 60cm 3 30cm 3 60cm 3 therefore 60 cm 3 of oxygen reacted with the 30 cm 3 of methane to produce 30 cm 3 of carbon dioxide

13. Molar volume If a volume of gas contains a fixed number of molecules, then it is useful to know what volume contains the Avogadro number of molecules. This volume is known as Molar volume. Molar volume depends on standard temperature of 0 o C (273K) and a pressure of 101.3 kPa. Under the above conditions, molar volume is taken as 24000 cm 3 or (24 dm 3 )mol -1. Main statement At room temperature and pressure ( r.t.p), 1 mol of a gas occupies 24000 cm 3 or 24dm 3 (Hint: you need to find the number of moles of the gas in order to calculate the molar volume at r.p.t, and where necessary use mole ratio from the balanced reaction equation)

14. EXAMPLES 1. 2.50g of calcium carbonate was heated and decomposed as in the equation below. CaCO 3 (s)  CaO(s) + CO 2 (g) (a) calculate the amount, in mol, of CaCO 3 that decomposed. n = m = 2.50 = 2.50 = 0.025 mol M {40 + 12 + (3x16)} 100 (b) calculate the amount, in mol, of carbon dioxide formed. Reaction equation: CaCO 3 (s)  CaO(s) + CO 2 (g) mole ratio: 1 mol: 1 mol: 1 mol amounts of mole required 0.025 0.025 0.025 i.e. 0.025 mol of carbon dioxide formed. (c) calculate the volume of carbon dioxide that formed. V = n x 24.0dm 3 = 0.025 x 24 = 0.60 dm 3

15. 2. Calculate the volume of hydrogen produced when excess dilute sulphuric acid is added to 5.00 g of zinc. Assume that under the conditions of the reaction 1 mol of gas has a volume of 24 dm 3. Answer: amount in moles of zinc used, n = m = 5.00 = 0.0765 mol M 65.4 the equation is: Zn(s) + H 2 SO 4 (aq)  ZnSO 4 (aq) + H 2 (g) mole ratio 1 mol 1 mol 1 mol 1 mol amounts required 0.0765 mol 0.0765 mol it follows that 0.0765 mol of hydrogen is produced molar volume = 24.0 dm 3 volume of hydrogen produced, V = n x 24.0 = 0.0765 x 24 = 1.836dm 3

16. 3. Calculate the volume of hydrogen produced when excess dilute sulphuric acid is added to 5.00 g of aluminium. Under the conditions of the reaction, 1 mol of gas has a volume of 24.0 dm 3. Answer amount of mole of Al used, n = m = 5.00 = 0.185 mol M 27.0 the equation for the reaction is: 2Al(s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 3H 2 (g) mole ratio 2 mol 3 mol 1 mol 3 mol simplest ratio 1 mol 1.5 mol 0.5 mol 1.5mol the equation indicates that 1 mol of aluminium produces 1.5 mol of hydrogen. Amount in moles of hydrogen produced = 1.5 x 0.185 = 0.278 mol hence, volume of hydrogen produced, V = n x 24.0 = 0.278 x 24 = 6.667 dm 3

17. 4. When reacted with excess dilute hydrochloric acid, what mass of calcium carbonate produces 1 dm 3 of carbon dioxide? Assume that 1 mol of of gas has a volume of 24 dm 3 Answer amount in moles of carbon dioxide = 1 = 0.0417 mol 24 the equation for the reaction is: CaCO 3 (s) + 2HCl(aq)  2CaCl(aq) + CO 2 (g) + H 2 O(l) mole ratio: 1 mol 2 mol 2 mol 1 mol 1 mol from the reaction equation 1 mol of carbon dioxide is produced by the reaction of 1 mol of calcium carbonate with excess hydrochloric acid. Therefore amount in moles of CaCO 3 used = 0.0417 mol mass of CaCO 3 used, m = n x M M(CaCO3) = 40. 0 + 12.0 + (3 x 16.0) = 100 hence, mass of CaCO 3 used = n x M = 0.0417 x 100 = 4.17 g

18. 5. A gas syring has a maximum capacity of 100 cm 3. Calculate the mass of copper(II) carbonate that would have to be heated to produce enough carbon dioxide to just fill the syringe at r.t.p Answer syringe volume of 100 cm 3 if equivalent to 100 = 0.00417 mol 24000 thus, volume of carbon dioxide to fill the syringe is 0.00417 mol the equation for the reaction is: CuCO 3 (s)  CuO(s) + CO 2 (g) mole ratio 1 mol 1 mol 1 mol amounts required 0.00417 mol 0.00417 mol thus, 0.00417mol of carbon dioxide is obtained by heating 0.00417mol of copper carbonate. Mass of copper carbonate, m = n x M = 0.00417 x ( 63.5 + 12 + 16x3) = 0.00417 x 123.5 = 0.515 g

19. 6. Calculate the volume of 8 g of methane at r.t.p Answer amount in mol of methane, CH 4, n = m = 8 = 8 M 12 + (4 x 1) 16 amount in mol of methane = 0.5 mol volume of 1 mol of methane at r.t.p = 24.0 dm 3 hence, volume of 0.5 mol of methane = 0.5 x 24 = 12 dm 3

20. Reacting mass, gas and solution volumes You will need to recall some of the formulae used earlier on and use them in such calculations. Rearrangement of the formulae may be necessary in some instances in order to do other calculations.

21. Recall 1. From mass: n = m or m = n x M M 2. From gas volumes: n = V(dm 3 ) or V(dm 3 ) = n x 24.0 24.0 3. From solutions: n = c x V(dm 3 ) or c = n V(dm 3 )

22. 1. A chemist reacted 0.23g of sodium with water to form 250 cm 3 of aqueous sodium hydroxide. Hydrogen gas was also produced. The equation is shown below. 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g) (a) calculate the amount, in mol, of Na that reacted. n = m = 0.23 = 0.010 mol M 23 (b) Calculate the volume of H 2 formed at room temperature and pressure. The equation for the reaction is 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g) mole ratio 2 mol 1 mol amounts required 0.010mol 0.005mol thus, 0.010 mol of Na produces 0.005 mol H 2 gas. Volume of hydrogen gas formed at r.t.p, V = n x 24000 cm 3 = 0.005 x 24000 = 120 cm 3 (c ) Calculate the concentration, in mol/dm 3, of NaOH(aq) formed. Reaction equation: 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g) mole ration: 2 mol 2 mol amount required 0.010 mol 0.010mol thus, 0.010 mol Na formed 0.010 mol of NaOH solution concentration of NaOH, in mol/dm 3, c = n = 0.010 mol = 0.04 mol/dm 3 V dm 3 0.250 dm 3

23. (aq) 2. Balance the equation below. MgCO 3 (s) + HNO 3 (aq)  Mg(NO 3 ) 2 (aq) + H 2 O(l) + CO 2 (g) (b) 2.529 g of MgCO 3 reacts with an excess of HNO 3. (i) what volume of CO 2, measured at RTP, is formed? (ii) The final volume of the solution is 50 cm 3. What is the concentration, in mol/dm 3, of Mg(NO 3 ) 2 formed

24. Now try questions 1 – 3 Page 21