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]

b) The mole ratio of aluminum hydroxide to sulfuric acid is 2:3. This means that every 1 mole of the aluminum hydroxide would require 1.5 moles of sulfuric acid.

0.5 mole aluminum hydroxide would require:

0.5 x 3/2 = 0.75 moles of sulfuric acid.

But only 0.500 moles of sulfuric acid is present. Thus, the limiting reagent is sulfuric acid.

c) With 0.5 moles sulfuric acid, the mole of aluminum hydroxide required would be:

0.5 x 2/3 = 0.33

Excess moles of aluminum hydroxide = 0.5 - 0.33

= 0.17 moles

d) The mole ratio of sulfuric acid to aluminum sulfate produced is 3:1. With 0.5 moles sulfuric acid, the mole of aluminum sulfate produced would be:

0.5 x 1/3 = 0.17 moles

But the percentage yield is 77%

77/100 x 0.17 = 0.13 moles

Thus, the moles of aluminum sulfate that would be obtained with a percentage yield of 77% would be 0.13 moles.

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