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0.500 mol aluminium hydroxide, Al(OH)3 reacts with 0.500 mol sulphuric acid, H2SO4 to produce aluminium sulphate and water.

a) Write the balanced equation for the reaction.

b) Which reactant is limiting reactant?

c) How many moles of excess reactant is used in the experiment?

d)Determine how many moles of aluminium sulphates was obtained if the percentage yield of aluminium sulphate during the experiment is 77%. [9 marks]​

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The number of moles of excess reactant is used in the experiment is 0.33 moles. The number of moles of aluminium sulphates was obtained  0.129 moles if the percentage yield is 77%.

2Al(OH)3(aq) + 3H2SO4(aq) -----> Al2(SO4)3(aq) + 6H2O(l)

Number of moles of Al(OH)3 = 0.500 mol

Number of moles of H2SO4 = 0.500 mol

We have to determine the limiting reactant as follows;

2 moles of Al(OH)3 reacts with 3 moles of H2SO4

0.500 moles of  Al(OH)3 reacts with 0.500 moles ×  3 moles/2 moles

= 0.75 moles of H2SO4.

This means that H2SO4 is limiting reactant

Since  3 moles of H2SO4 reacts with 2 moles of Al(OH)3

0.500 moles of  H2SO4 reacts with  0.500 × 2 moles /3 moles

= 0.33 moles

Only  0.33 moles of excess reactant was used in the experiment.

3 moles of  H2SO4 yields 1 mole of  Al2(SO4)3

0.500 moles of  H2SO4 yields  0.500 moles ×  1 mole/3 moles = 0.167 moles

Theoretical yield of Al2(SO4)3 = 0.167 moles

% yield = actual yield/Theoretical yield × 100/1

Actual yield = % yield × Theoretical yield /100

Actual yield =  77 ×  0.167/100

Actual yield = 0.129 moles

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