Answer to Question #344933 in Chemistry for Aicé

Question

A mixture of Al and Fe weighing 0.2000 g is treated with excess HCI solutions. Each metal reacts separately with the acid, resulting in the production of hydrogen gas H2, and a chloride salt. After treatment, a total volume of 200mL of the evolved gas was collected at 20 °C and 850 torr. Calculate for the mass percentage of Al in the sample.

Answer 1

Solution:

According to the ideal gas law: PV = nRT

P = 850 torr

V = 200 mL = 0.2 L

R = 62.36 L torr mol⁻¹ K⁻¹

T = 20°C + 273.15 = 293.15 K

So,

n(H₂) = PV / RT = (850 torr × 0.2 L) / (62.36 L torr mol⁻¹ K⁻¹ × 293.15 K) = 0.0093 mol H₂

The molar mass of Al is 26.98 g/mol

The molar mass of Fe is 55.845 g/mol

The given reactions are:

(1): 2Al + 6HCl → 2AlCl₃ + 3H₂

(2): Fe + 2HCl → FeCl₂ + H₂

Consider, mass of Al as X and mass of Fe as Y

As the total weight of Al and Fe is 0.2000 g

Thus,

X + Y = 0.2000

According to the reaction (1):

n₁(H₂) = (X × 3) / (2 × 26.98) = 0.0556X

According to the reaction (2):

n₂(H₂) = (Y × 1) / (1 × 55.845) = 0.0179Y

As the total number of moles of H₂ is 0.0093 mol

Thus,

0.0556X + 0.0179Y = 0.0093

1) X + Y = 0.2000

2) 0.0556X + 0.0179Y = 0.0093

Multiplying the first equation by 0.0179 and subtracting the second equation gives:

−0.0377X = −0.00572

X = 0.1517

Mass of Al = X = 0.1517 g

%Al = (Mass of Al / Mass of mixture) × 100%

%Al = (0.1517 g / 0.2000 g) × 100% = 75.85%

%Al = 75.85%

Answer: The mass percentage of Al in the sample is 75.85%