1. Ready for learning? Bags to the side / back Drinks and food away Badges on please 1 Learning aims:  How to work out the products of a chemical reaction.  To complete new material – reacting masses  To prepare for the theory exam next week  Carry out practise of practical for practical exam in two weeks. News !

2. Periodic table  There are lots of different types of reactions:  We will look at some briefly here.  The only one you will need to know for the exam is the acid-base reaction  We will come back to others later in the course. 2

3. Periodic table  Direct Combination or Synthesis Reaction In a synthesis reaction two or more chemical species combine to form a more complex product. A + B → AB  8 Fe + S 8 → 8 FeS  Chemical Decomposition In a decomposition reaction a compound is broken into smaller chemical species. AB → A + B The electrolysis of water into oxygen and hydrogen gas is an example of a decomposition reaction: 2 H 2 O → 2 H 2 + O 2

4. Periodic table  Single Displacement or Substitution Reaction A substitution or single displacement reaction is characterized by one element being displaced from a compound by another element. A + BC → AC + B e.g. zinc combines with hydrochloric acid, the zinc replaces the hydrogen: Zn + 2 HCl → ZnCl 2 + H 2  Double Displacement Reaction In a double displacement or metathesis reaction two compounds exchange bonds or ions in order to form different compounds. AB + CD → AD + CB E.g. sodium chloride and silver nitrate to form sodium nitrate and silver chloride. NaCl(aq) + AgNO 3 (aq) → NaNO 3 (aq) + AgCl(s)

5. Periodic table  Oxidation-Reduction or Redox Reaction  Oxidation Is Loss Reduction Is Gain Redox reactions involve the transfer of electrons between chemical species. The reaction that occurs when In which Cl 2 is reduced to Cl - and KBr (bromide anion) is oxidized to bromine provides an example of a redox reaction: KBr(aq) + Cl 2 (aq) → Br 2 (aq) + 2KCl(aq)  A combustion reaction - oxygen combines with another compound to form carbon dioxide and water.  E.g. the burning of naphthalene: C 10 H 8 + 12 O 2 → 10 CO 2 + 4 H 2 O

6. Periodic table  An acid-base reaction is type of double displacement reaction that occurs between an acid and a base. The H + ion in the acid reacts with the OH - ion in the base to form water and an ionic salt: HA + BOH → H 2 O + BA The reaction between hydrochloric acid (HCl) and sodium hydroxide is an example of an acid-base reaction: HCl + NaOH → NaCl + H 2 O

7. Periodic table Acid + base gives salt and water HA + BOH  BA + H 2 O Let’s try a few: 7

8. Periodic table 8

9.  We have learnt how to work out the Mr of a compound.  We have learnt how to calculate the mass of a compound by using m = Mr n  And to calculate the number of moles n = m/Mr  We have used the titration formula to calculate an unknown concentration or volume. M A V A = x  M B V B y 9

10. Periodic table  Chemists often need to calculate the amount of a particular product that will be obtained from a certain amount of reactant.  Or the amount of reactant required in order to obtain a certain amount of product.  We always start this calculation off with a balance equation. 10

11. Periodic table 11 12 g of magnesium burns in air. Calculate 2Mg + O 2 12 g 2MgO the mass of magnesium oxide formed ? g STAGE 1 convert mass of Mg into moles of Mg use the equation to convert moles of Mg into moles of MgO STAGE 3 convert moles of MgO into mass of MgO moles of Mg moles of MgO STAGE 2 n=m/Mr m =Mr n

12. Periodic table 12 STAGE 1 STAGE 2 from the equation n(Mg) = n(MgO) STAGE 3 Mr(MgO) = 40 n(MgO) = 0.5 mol m(MgO) = 0.5 x 40 24 12 n(Mg) = = 24Mr(Mg) = 0.5 mol = 20 g n=m/Mr m =Mr n

13. Periodic table 13 What mass of barium sulphate would be produced from 10 g of barium chloride in the following reaction? BaCl 2 + H 2 SO 4  BaSO 4 + 2HCl 10 g ? M r (BaCl 2 ) = 208 M r (BaSO 4 ) = 233 10/208 = n(BaSO 4 ) = m(BaSO 4 ) = n(BaCl 2 ) = 0.048 mol 11.184 g 0.048 mol 0.048 x 233 = n=m/Mr m =Mr n

14. Periodic table 14 What mass of calcium carbonate is needed to remove 10.0 g of SO 2 in the following reaction? 2CaCO 3 (s) + 2SO 2 (g) + O 2  2CaSO 4 (s) + 2CO 2 (g ) ?10 g M r (SO 2 ) = 64 M r (CaCO 3 ) = 100 10/64 = n(CaCO 3 ) = m(CaCO 3 ) = n(SO 2 ) = 0.156 mol 15.6 g 0.156 mol 0.156 x 100 = Note! Even though we have 2 : 2 ratio we can still deal with 1 amount to 1 amount because they are in the same proportions n=m/Mr m =Mr n

15. Periodic table 15 What mass of KIO 3 is required to give 10 g of iodine in the following reaction? KIO 3 (aq) + KI(aq) + 6H + (aq)  3I 2 (aq) + 6K + (aq) + 3H 2 O(l) ?10 g M r (I 2 ) = 254 M r (KIO 3 ) = 214 10/254 = n(KIO 3 ) = m(KIO 3 ) = n(I 2 ) = 0.0394/3 = 2.81 g 0.0394 mol 0.0131 x 214 = 0.0131mol This time we have a 1:3 ratio, so for every 10g of iodine produced we need 1/3 of the amount of KIO 3 n=m/Mr m =Mr n

16. Periodic table 16 What mass of sodium carbonate can be obtained by heating 100 g of sodium hydrogencarbonate? 2NaHCO 3 (s)  Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) 100 g? M r (NaHCO 3 ) = 84 M r (Na 2 CO 3 ) = 106 100/84 = n(Na 2 CO 3 ) = m(Na 2 CO 3 ) = n(NaHCO 3 ) = 1.19/2 = 63.07 g 1.19 mol 0.595 x 106 = 0.595 mol here we have 2:1 ratio so we will need ½ the amount of sodium carbonate.

17. Periodic table 17 What mass of sodium chloride is required to produce 1 kg of sodium hydrogencarbonate in the following reaction? NaCl(aq) + NH 3 (g) + H 2 O(l) + CO 2 (g)  NaHCO 3 (s) + NH 4 Cl(aq) ?1000 g M r (NaHCO 3 ) = 84 M r (NaCl) = 58.5 1000/84 = n(NaCl) = m(NaCl) = n(NaHCO 3 ) = 11.90 mol 696.15 g 11.90 mol 11.90 x 58.5 =

18. Periodic table 18 What mass of ethyne can be prepared from 10.0 g of calcium carbide by the following reaction? CaC 2 (s) + 2H 2 O(l)  Ca(OH) 2 (aq) + C 2 H 2 (g)? 10 g? M r (CaC 2 ) = 64 M r (C 2 H 2 ) = 26 10/64 = n(C 2 H 2 ) = m(C 2 H 2 ) = n(CaC 2 ) = 0.156 mol 4.056 g 0.156 mol 0.156 x 26 =

19. Periodic table 19 Calculate the mass of ammonium chloride required to produce 1 kg of ammonia 2NH 4 Cl(s) + Ca(OH) 2  2NH 3 (g) + CaCl 2 (s) + 2H 2 O(g) ?1 kg M r (NH 4 Cl) = 53.5 M r (NH 3 ) = 17 1000/17 = n(NH 4 Cl) = m(NH 4 Cl) = n(NH 3 ) = 58.82 mol 3.15 kg 58.82 mol 58.82 x 53.5 =

20. Periodic table 20 What mass of potassium chloride would be produced from the reaction of 20 g of potassium carbonate with excess hydrochloric acid? K 2 CO 3 (s) + 2HCl(aq)  2KCl(aq) + CO 2 (g) + H 2 O(l) 20 g? M r (K 2 CO 3 ) = 138 M r (KCl) = 74.5 20/138 = n(KCl) = m(KCl) = n(K 2 CO 3 ) = 2 x 0.145 = 21.61 g 0.145 mol 0.290 x 74.5 = 0.290 mol

21. Periodic table 21

22. Periodic table  This is a solution of known concentration.  This will be 250cm 3 0.1M solution of sodium hydroxide  We need to know how many grams to weigh out.  Use the equation n = MV 1000 moles = 0.1 x 250 / 1000 = 0.025 Convert moles to mass …….m = Mr n mass = (23+16+1) x 0.025 = 1g 22

23. 23 Periodic Table click to return