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Sulfuric acid reacts with aluminum hydroxide by double replacement.
a. if 30.0 g of sulfuric acid reacts with 25.0 g of aluminum hydroxide, identify the limiting reactant.
b. determine the mass of excess reactant remaining.
c. determine the mass of each product formed. assume 100% yield.

Respuesta :

Ishankahps

a. 
The balanced equation for the reaction between sulfuric acid and aluminium hydroxide is,
                        3H₂SO₄     +   2Al(OH)₃ → Al₂(SO₄)₃ +   6H₂O

mass of H₂SO₄  =        30.0 g             
Molar mass of H₂SO₄  =   98 g/mol
moles of H₂SO₄ = 30.0 g /98g /mol = 0.306 mol

mass of Al(OH)₃            =         25.0 g            
Molar mass of Al(OH)₃  =   78 g/mol
moles of Al(OH)₃           = 25.0 g/ 78 g/mol = 0.321 mol

Stoichiometric ratio between H₂SO₄  and Al(OH)₃ is 3 : 2

Hence reacted moles of H₂SO₄ = 0.306 mol
            reacted moles of Al(OH)₃ = 0.306 mol x (2 / 3) = 0.204 mol

Hence the limiting reactant is H₂SO₄ 

b.
According to the above calculation, the excess reactant is Al(OH)₃. 

The reacted moles of Al(OH)₃ = 0.306 mol x (2 / 3) = 0.204 mol

The added moles of Al(OH)₃ = 0.321 mol

Hence the remaining Al(OH)₃ moles = added moles - reacted moles
                                                          = 0.321 mol - 0.204 mol
                                                          = 0.117 mol

Molar mass of Al(OH)₃  =   78 g/mol
Remaining mass of Al(OH)₃ = number of moles x molar mass
                                             = 0.117 mol x 78 g/mol
                                             = 9.126 g

c. 

The products formed from the reaction between aluminium hydroxide and sulfuric acid are Al₂(SO₄)₃ and H₂O

The limiting reactant is H₂SO₄ 

The stoichiometric ratio between H₂SO₄  and Al₂(SO₄)₃ is 3 : 1
Reacted moles of H₂SO₄ = 0.306 mol
Hence the moles of Al₂(SO₄)₃ formed = 0.306 mol / 3
                                                             = 0.102 mol
Molar mass of Al₂(SO₄)₃  = 342 g/mol
Mass of Al₂(SO₄)₃  formed = 0.102 mol x 342 g/mol
                                           = 34.884 g

The stoichiometric ratio between H₂SO₄  and H₂O is 3 : 6
Reacted moles of H₂SO₄ = 0.306 mol
Hence the moles of H₂O formed = 0.306 mol x (6 / 3)
                                                    = 0.612 mol
Molar mass of H₂O  = 18 g/mol
Mass of H₂O  formed = 0.612 mol x 18 g/mol
                                   = 11.016 g


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