If 30.0 grams of sulfuric acid react with 26.0 grams of aluminum hydroxide in a double replacement reaction, how many grams of water can be produced? How many grams of the excess reactant will be left over at the end of the reaction? (Don’t forget to balance the reaction!) (8 points) H2SO4 + Al(OH)3 → H2O + Al2(SO4)3

a) mass of water produced= 11.16 g of water

b) mass of excess reactant= 9.36 g of aluminum hydroxide

Explanation:

The balanced reaction equation is;

3H2SO4 + 2Al(OH)3 → 6H2O + Al2(SO4)3

The next step is to determine the limiting reactant. The limiting reactant will give the least number of moles of product.

For sulphuric acid;

Molar mass of sulphuric acid =98gmol-1

Number of moles of sulphuric acid= 30g/98gmol-1 = 0.31 moles of sulphuric acid

From the reaction equation;

3 moles of sulphuric acid yields 6 moles of water

0.31 moles of sulphuric acid will yield 0.31 ×6/3 = 0.62 moles of water

For aluminum hydroxide;

Number of moles= 26.0 g/78 g/mol = 0.33 moles

From the reaction equation;

2 moles of aluminum hydroxide yields 6 moles of water

0.33 moles of aluminum hydroxide will yield 0.33 × 6/2 = 0.99 moles of water

Hence sulphuric acid is the limiting reactant.

Thus mass of water produced= 0.62 moles ×18gmol-1 = 11.16 g of water

Since 3 moles of sulphuric acid reacts with 2 moles of aluminum hydroxide

0.31 moles of sulphuric acid reacts with 0.31 ×2/3 = 0.21 moles of aluminum hydroxide

Thus amount of excess reactant = 0.33- 0.21 = 0.12 moles of aluminum hydroxide

Mass of excess aluminum hydroxide = 0.12 × 78 g/mol = 9.36 g of aluminum hydroxide