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Understanding Chemical Reactions

Understanding Chemical Reactions. Lesson: Calculations in Chemistry 2. Page 175 Q 4. One mole of ZnO has a RAM of 65 + 16 = 81g 0.2 moles will have a mass of 0.2 X 81 = 16.2 g. Page 175 Q 4. b) 1 mole of H 2 S has a mass of (2x1) + 32 = 34g 2.5 moles will have a mass of

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Understanding Chemical Reactions

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  1. Understanding Chemical Reactions Lesson: Calculations in Chemistry 2

  2. Page 175 Q 4 • One mole of ZnO has a RAM of 65 + 16 = 81g 0.2 moles will have a mass of 0.2 X 81 = 16.2 g

  3. Page 175 Q 4 b) 1 mole of H2S has a mass of (2x1) + 32 = 34g 2.5 moles will have a mass of 2.5 X 34 = 85 g

  4. Page 175 Q 4 c) 1 mole of CuSO4 has a mass of 63.5 + 32 + (16 x 4) = 259.5g 0.45 moles will have a mass of 0.45 X 259.5 = 116.8 g

  5. Calculating formulae Page 178 of Chemistry text

  6. Working out the formula of magnesium oxide • A student heated some Mg as shown. • When Mg burns in air it combines with oxygen to make magnesium oxide.

  7. Working out the formula of magnesium oxide Step 1 - Results • Mass of crucible + lid + Mg before heating = 25.24g • Mass of crucible and lid = 25.00g • Therefore, mass of Mg = 0.24g

  8. Working out the formula of magnesium oxide Step 2 - Results • Mass of crucible + lid + magnesium oxide after heating = 25.40g • Mass of crucible + lid + Mg before heating= 25.24g • Therefore, mass of oxygen in magnesium oxide = 0.16g

  9. Working out the formula of magnesium oxide Step 3 – Change the masses into moles Magnesium = 0.24 / 24 = 0.01 mole Oxygen = 0.16 / 16 = 0.01 mole

  10. Working out the formula of magnesium oxide Step 4 – Work out the ratio of moles Mg : O 0.01 : 0.01 1 : 1 Therefore the formula of magnesium oxide is MgO

  11. One for you! • A compound of nitrogen and hydrogen was broken down into its elements. • It was found that 1.4g of nitrogen had combined with 0.3g of hydrogen in the compound. • What was the formula of the compound? • (R.A.M.s N = 14, H = 1)

  12. One for you! Step 1 – work out the number of moles Moles of N = 1.4 / 14 = 0.1 Moles of H = 0.3 / 1 = 0.3

  13. One for you! Step 2 – work out the ratio of the number of moles to the lowest whole numbers N : H 0.1 : 0.3 1 : 3 Therefore there is 3 times as many H atoms as N atoms. Its formula must be NH3

  14. Reacting iron with copper sulphate • A more reactive metal will displace a less reactive metal from its solution: • Iron + copper sulphate iron sulphate + copper • Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s) • How much copper will 0.5g of iron produce?

  15. Reacting iron with copper sulphate • Fill your beaker with copper sulphate solution. • Add precisely 5g of iron to the beaker and stir gently for 4 minutes. • Filter the mixture – making sure all the solid is removed from the beaker – rinse beaker with water if needed.

  16. Reacting iron with copper sulphate • Rinse the filter paper with propanone to dry the copper. • When dry – re-weigh the solid.

  17. 1 mole of Fe should produce 1 mole of Cu • 5 g of Fe is 5 / 56 = 0.09 moles • 0.09 moles of Fe should produce 0.09 moles of Cu • 0.09 moles of Cu will have a mass of 0.09 x 64 = 5.7 g

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