Question

You can dissolve an aluminum soft drink can in an aqueous base as potassium hydroxide.

2 Al + 2 KOH + 6 H2O = 2 KAl(OH)4 + 3 H2

If you place 2.05g of aluminum in a beaker with 185ml of 1.35 M KOH, will any aluminum remain? What mass of KAl(OH)4 is produced?

Answer 1

m = 2.05 g of Aluminium

mol = mass/MW = 2.05/26.98 = 0.07598 mol of Al

mol of KOH = M*V = (1.35)(0.185) =0.24975 mol of KOH

since ratio is 2:2; that is 1:1

then

0.07598 mol of Al react with 0.07598 mol ofKOH

a)

no aluminium remains, aluminium is limiting reactant

b)

ratio between Al and KAl(OH)4

2:2

so

1:1

therefore

0.07598 mol of Al react to form 0.07598 mol of KAl(OH)4

MW of KAl(OH)4= 147.02445 g/mol

then

mass = mol*MW = 0.07598*147.02445 = 11.170 g of KAl(OH)4 areproduced