Answer to Question #190027 in Chemistry for Caylin Adams

Question #190027

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 .


1
Expert's answer
2021-05-08T23:33:48-0400

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%.

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