Answer to Question #190027 in Chemistry for Caylin Adams

Question

Good day

My question is:

A mixture containing only FeCl₃ and AlCl₃ weighs 5.95 g. The chlorides are converted to hydrous oxides and ignited to Fe₂O₃ and Al₂O₃. The oxide mixture weighs 2.62 g.

Determine the percentage Fe and Al in the original mixture

please assist .

Answer 1

Molecular weight of FeCl₃ is 162.2 g/mol

Molecular weight of AlCl₃ is 133.34 g/mol

Molecular weight of Fe₂O₃ is 159.69 g/mol

Molecular weight of Al₂O₃ is 101.96 g/mol

Atomic weight Fe is 55.845 g/mol

Atomic weight Al is 26.98 g/mol

Solution:

Let, mass of Fe = a, mass of Al = b.

(1): FeCl₃ → Fe

(2): AlCl₃ → Al

According to the equation (1): g FeCl₃ = g Fe × (Molecular weight of FeCl₃ / Atomic weight Fe) × (1/1)

According to the equation (2): g AlCl₃ = g Al × (Molecular weight of AlCl₃ / Atomic weight Al) × (1/1)

Here, mass of FeCl₃ + AlCl₃ = 5.95 g

Hence,

g FeCl₃ + g AlCl₃ = 5.95

g Fe × (162.2 / 55.845) × (1/1) + g Al × (133.34 / 26.98) × (1/1) = 5.95

(g Fe × 2.9045) + (g Al × 4.9422) = 5.95

2.9045a + 4.9422b = 5.95

(3): Fe₂O₃ → 2Fe

(4): Al₂O₃ → 2Al

According to the equation (3): g Fe₂O₃ = g Fe × (Molecular weight of Fe₂O₃ / Atomic weight Fe) × (1/2)

According to the equation (4): g Al₂O₃ = g Al × (Molecular weight of Al₂O₃ / Atomic weight Al) × (1/2)

Here, mass of Fe₂O₃ + Al₂O₃ = 2.62 g

Hence,

g Fe₂O₃ + g Al₂O₃ = 2.62

g Fe × (159.69 / 55.845) × (1/2) + g Al × (101.96 / 26.98) × (1/2) = 2.62

(g Fe × 1.4298) + (g Al × 1.8895) = 2.62

1.4298a + 1.8895b = 2.62

Let's consider the two equations:

(I): 2.9045a + 4.9422b = 5.95

(II): 1.4298a + 1.8895b = 2.62

1) Multiply the equation (I) by -1.4298:

(2.9045a + 4.9422b = 5.95) × (-1.4298) = -4.1529a - 7.0664b = -8.5073

2) Multiply the equation (II) by 2.9045:

(1.4298a + 1.8895b = 2.62) × (2.9045) = 4.1529a + 5.4881b = 7.6098

3) Add the equation (I) to the equation (II), we have:

-4.1529a - 7.0664b = -8.5073

4.1529a + 5.4881b = 7.6098

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

-1.5783b = -0.8975

b = 0.56865 = 0.57

Mass of Al = b = 0.57 g

2.9045a + 4.9422b = 5.95

2.9045a + 4.9422 × 0.56865 = 5.95

2.9045a = 3.1396

a = 1.08095 = 1.08

Mass of Fe = a = 1.08 g

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

%Fe = (1.08 g / 5.95 g) × 100% = 18.15%

%Fe = 18.15%

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

%Al = (0.57 g / 5.95 g) × 100% = 9.58%

%Al = 9.58%

Answer: %Fe = 18.15%; %Al = 9.58%.