Question

A 0.250-g sample of a magnesium-aluminum alloy dissolves completely in an excess of HCl(aq). When the liberated H2(g) is collected over water at 29 ∘C and 752 torr, the volume is found to be 343 mL . The vapor pressure of water at 29 ∘C is 30.0 torr.

What is the mass percentage of aluminum in this alloy?

Express your answer numerically.

Answer 1

Ans:

Temperature, T = 29^oC = 302 K

Volume of H2 when collected over water, V = 343 mL = 343 x 10^-6 m³

Pressure due to H2 and water vapour = 752 torr

Vapour pressure of water at 302 K = 30.0 torr

The pressure of H2 gas = 752 - 30 = 722 torr = 96258.48Pa = P (1 torr = 133.322 Pa)

No. of moles of H2 gas(n) collected can be obtained by ideal gas equation;

PV = nRT

where, R = gas constant = R = gas constant = 8.314 m³⋅Pa⋅K⁻¹⋅mol⁻¹

n = (PV)/(RT) = (96258.48Pa x 343 x 10^-6 m³)/( 8.314 m³⋅Pa⋅K⁻¹⋅mol⁻¹ x 302 K) = 0.01315 mol

No. of moles of H2 gas(n) collected = 0.01315 mol

Mass of alloy = 0.250 g

Let ‘xg’ be the mass of aluminium in the alloy.

Then mass of Mg in the alloy = (0.250-x)g’.

No. of moles of Mg = [(0.250-x)/(24.305)] mol     {atomic mass of Mg = 24.305 g/mol}

No. of moles of aluminium = (x/26.98) mol    { atomic mass of Al = 26.98 g/mol}

Reactions:

Mg + 2HCl ----> MgCl2 + H2    (1 mol of Mg releases 1 mol of H2 gas)

1Al + 3HCl ----> 1AlCl3 + 1.5 H2   (1 mol of Al releases 1.5 mol of H2 gas)

[(0.250-x)/(24.305)] + 1.5 (x/26.98) = 0.01315

Solving for ‘x’, we get, x = 0.1982 g = mass of aluminium

Mass % of aluminium = {(mass of aluminium)/(total mass of alloy)} x 100

Mass % of aluminium = (0.1982/0.250) x 100 =79.28%