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Higher Chemistry Unit 1(a) Identifying a reactant in excess.

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1 Higher Chemistry Unit 1(a) Identifying a reactant in excess

2 After today’s lesson you should be able to: Use balanced equations, n = c x v and n = m/gfm to calculate which reactant is in excess. Use the reactant which is not in excess to calculate the mass of the product.

3 Reactants in excess In order to ensure a reaction goes to completion, too much of one reactant is added to ensure all of the other reactant(s) is used up. The reactant of which there is too much of is said to be ‘in excess’ and must be identified if the mass of product has to be calculated.

4 Example 1 0.6g of magnesium ribbon was added to 40cm 3 of 2mol l -1 hydrochloric acid. (a)Show by calculation which reactant is in excess (b)Calculate the mass of salt produced.

5 (a)Step 1: Write a balanced equation Mg + 2HCl → MgCl 2 + H 2

6 (a)Step 2: Calculate the mole to mole relationship of the reactants Mg + 2HCl → MgCl 2 + H 2 1 mol 2 mols

7 (a)Step 3: Calculate the actual number of moles of each reactant Mg: 0.6g n = m/gfm = 0.6/24.3 = 0.025mol HCl: 40cm 3 of 2mol l -1 n = c x v = 2 x 0.04 = 0.08mol

8 (a)Step 4: Compare actual number of moles present to mole to mole relationship From balanced equation: 1 mol Mg reacts with 2 mols HCl So from actual number of mol present: 0.025mol Mg would react with 0.05mol HCl BUT 0.08mol present therefore HCl in excess by 0.03mol.

9 (b) Step 1: work out the mol to mol relationship Using the reactant which is NOT in excess Mg + 2HCl → MgCl 2 + H 2 1 mol 1 mol So: 0.025mol 0.025mol

10 (b) Calculate the mass of salt m = n x gfm = x 95.3 = 2.38g

11 Example 2 2.6g of zinc powder was added to 30cm 3 of 1mol l -1 copper(II) sulphate solution. (a)Show by calculation which reactant is in excess (b)Calculate the mass of copper produced.

12 (a)Step 1: Write a balanced equation Zn + CuSO 4 → ZnSO 4 + Cu

13 (a)Step 2: Calculate the mole to mole relationship of the reactants Zn + CuSO 4 → ZnSO 4 + Cu 1 mol 1mol

14 (a)Step 3: Calculate the actual number of moles of each reactant Zn: 2.6g n = m/gfm = 2.6/65.4 = 0.04mol CuSO 4 : 30cm 3 of 1mol l -1 n = c x v = 1 x 0.03 = 0.03mol

15 (a)Step 4: Compare actual number of moles present to mole to mole relationship From balanced equation: 1 mol Zn reacts with 1mol CuSO 4 So from actual number of mol present: 0.04mol Zn would react with 0.04mol CuSO 4 BUT Only 0.03mol CuSO 4 present therefore Zn in excess.

16 (b) Step 1: work out the mol to mol relationship Using the reactant which is NOT in excess Zn + CuSO 4 → ZnSO 4 + Cu 1 mol 1 mol So: 0.03mol 0.03mol

17 (b) Calculate the mass of Cu m = n x gfm = 0.03 x 63.5 = 1.905g


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