Answer:
There will remain 11.47 grams of Al(OH)3
Explanation:
Step 1: Data given
Mass of sulfuric acid = 35.0 grams
Molar mass sulfuric acid = 98.08 g/mol
Mass of aluminium hydroxide = 30.0 grams
Molar mass of aluminium hydroxide = 78.0 g/mol
Step 2: The balanced equation
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles Al(OH)3 = 30.0 grams / 78.0 g/mol
Moles Al(OH)3 = 0.385 moles
Moles H2SO4 = 35.0 grams / 98.08 g/mol
Moles H2SO4 = 0.357 moles
Step 4: Calculate the limiting reactant
For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20
H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles
There will remain 0.385 - 0.238 = 0.147 moles
Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams
There will remain 11.47 grams of Al(OH)3
