Question

An alloy of aluminum and magnesium was treated with sodium hydroxide solution, in which only aluminum reacts.

2Al(s) + 2NaOH(aq) + 6H₂O(l)----> 2NaAl(OH)₄(aq) +3H₂(g)

If a sample of alloy weighing 1.225g gave 0.1093g of hydrogen, what is the percentage of aluminum in the alloy?

Answer 1

Answer

79.61%

Explanation

2Al(s) + 2NaOH + 6H2O ---------> 2NaAl(OH)4(aq) + 3H2(g)

Stoichiometrically, 2moles of Al gives 3moles of H2

Number of moles = mass/molar mass

Moles of H2 produced = 0.1093g/2.016g/mol = 0.054216mol

moles of Al = (2/3) × 0.054216mol = 0.036144mol

Mass of Al = 0.036144mol × 26.981g/mol = 0.9752g

% Aluminium = (0.9752g/1.225g) × 100 = 79.61%