Question

Magnesium-aluminum alloys are commonly used in aircraft construction. When

treated with acid, both metals react to form hydrogen gas, as shown in the unbalanced

chemical equations below. A 1.000 g sample of alloy produces 0.107 g of H2 (g) when reacted

with excess HCl (aq). What is the percent composition by mass of Al in the alloy?

Al (s) + HCl (aq) à AlCl3 (aq) + H2 (g)

Mg (s) + HCl (aq) à MgCl2 (aq) + H2 (g)

Answer 1

2Al + 6HCl ------> 2AlCl3 + 3H2 .....(1)

Mg + 2HCl -----> MgCl2 + H2 ......(2)

Let mass of Al in alloy = (x) g

Moles of Al = mass/molar mass = x/27 moles

Then, mass of Mg in alloy = (1 - x) g

Moles of Mg = mass/molar mass = (1 - x)/24 moles

From reaction;

2 mole Al produces 3 moles H2

So, x/27 mole Al will produce = 1.5 * (x/27) moles of H2 = 1.5x/27 moles

Moles of H2 = 1.5x/27

From reaction (2)

1 moles Mg produces 1 mole H2

So, (1-x)/24 mole Mg will produce = (1-x)/24 moles of H2

Moles of H2 = (1 - x)/24

Total moles of H2 = (1.5x/27) + ((1 - x)/24)

= (1.5x*24 + 27 - 27x)/(27*24) = (9x + 27)/648 moles = (x+3)/72

Total mass of H2 = moles * molar mass = ((x+3)/72)*2 = (x+3)/36 g

So,

x+3/36 = 0.107

x = 0.852 grams

So,

% of Al in x = 0.852*100/1 = 85.2 % ..... Answer