Answer:
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
