Question

In the reaction: 3H2SO4 + 2Al(OH)3 → Al2 (SO4)3 + 6H2O,

If 25.0 g of sulfuric acid reacts with 20.8 g of aluminum hydroxide,

a) Which is the limiting reactant?

b) Determine the mass of excess reactant remaining.

c) Determine the mass of each product formed.

Answer 1

Moles of sulfuric acid = $\frac{25}{98.079}=0.25$

Moles of aluminium hydroxide = $\frac{20.8}{78}=0.27$

limiting reactant is aluminium hydroxide

$Moles [H_20]$ = 3 × 0.25 = 0.75

Mass = 0.75 × 18 =13.5 grams