Question

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.

Answer 1

To calculate the percent composition of aluminum and magnesium in the sample, we use stoichiometry. Moles of hydrogen gas produced help to establish moles of the metals reacting, but without complete reaction equations for the alloy, or more data, a definitive calculation cannot be provided.

Step-by-step explanation:

To calculate the percentage by mass of aluminum and magnesium in the sample, we need to perform a few stoichiometric calculations based on the reaction of the alloy with hydrochloric acid and the production of hydrogen gas. The typical reactions for aluminum and magnesium with hydrochloric acid are:

• 2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g)
• Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)

Using the molar mass of hydrogen (1.008 g/mol), we can calculate the moles of hydrogen gas produced using the equation:

moles H₂ = 0.998 g / (2 * 1.008 g/mol) = 0.495 moles

The mole ratio of Al to H₂ in the reaction is 2:3, and for Mg to H₂ is 1:1. We can calculate the mass of aluminum or magnesium that would produce 0.495 moles of hydrogen.

For aluminum:

moles Al = (2/3) * moles H₂ = (2/3) * 0.495 = 0.330 moles

For magnesium:

moles Mg = moles H₂ = 0.495 moles

Now, we calculate the mass:

mass Al = moles Al * atomic mass Al = 0.330 moles * 26.98 g/mol = 8.904 g

mass Mg = moles Mg * atomic mass Mg = 0.495 moles * 24.31 g/mol = 12.033 g

These numbers are hypothetical maximums if the sample was 100% Al or Mg, and do not add up to 9.87 g. Therefore, we need to set up a system of equations considering the total mass of the alloy and the mass of hydrogen produced to find the exact mass of Al and Mg in the sample. This requires more information or a different approach that accounts for the actual reaction stoichiometry with the alloy sample, which isn't provided here.

Answer 2

The answer is

The percentage by mass of Aluminium is 62.479% Al

The percentage by mass of Magnesium is 37.52% Mg

Step-by-step 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