A 9.87-gram sample of an alloy of aluminum and magnesium is completely reacted with hydrochloric acid and yields 0.998 grams of hydrogen gas. Calculate the percentage by man of each metal in the alloy.

The answer is

The percentage by mass of Aluminium is 62.479% Al

The percentage by mass of Magnesium is 37.52% Mg

Explanation:

The solution to this can be found by listing out the known variables

mass of alloy = 9.87 grams

Mass of hydrogen gas produced = 0.998 grams

Molar mass of Al = 26.98 g/mol

Molar mass of Mg = 24.3 g/mol

Molar mass of H₂ (g) = 2.016 g/mol

The reaction between aluminium, magnesium and HCl aare as follows

2Al (s) + 6HCl (aq) → 2AlCl₃ (aq) + 3H₂ (g)

Mg (s) + 2HCl (aq) → MgCl₂ (aq) + H₂ (g)

From the quation for the reactions, it is seen that 2 moles of Al produces 3 moles of H₂ (g) thus if the mass of aluminium in the alloy is say x we have

2*x/26.98 moles of Aluminium produces 3*x/26.98 moles of H₂

or one mole produces 0.0555967x moles H₂ (g)

Also 1 mole of magnesium produces 1 mole of H₂ (g)

hence since the mass of magnesium in the sample is (9.87 - x) grams, we have

(9.87-x) / 24.3 moles of Mg produces (9.87 - x) / 24.3 moles of H₂ (g)

However the number of moles of H₂ actually produced = 0.998/2.016 = 0.495 moles H₂

Therefore 0.0555967x moles + (9.87 - x) / 24.3 moles = 0.495 moles

or 0.0555967x + 0.4062 - 0.04115x = 0.495

0.0144x = 0.0888, Terefore x = 6.16 grams

and the percentage by mass of Al in the alloy is

6.16/9.87 * 100 = 62.479% Al and

(100 - 62.47) or 37.52% Mg

The percentage by mass of Aluminium is 62.479% Al

The percentage by mass of Magnesium is 37.52% Mg