Question

Over the years, the thermite reaction has been used for years to welding railroad nails, in cendiary bombs, and to ignite solid fueled rocket motors. The reaction is

Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s)
A. What mass of iron(III) oxide must be used to produce 76.73 g of iron?
B. What mass of aluminum must be used to produce 76.72 g of iron?
C. What is the maximum mass of aluminum oxide that could be produced along with 76.75 g iron?

Answer 1

A. 109.61 g of Fe₂O₃

B. 36.99 g of Al

C. Maximum mass of Al₂O₃ produced is 69.90 g

Step-by-step explanation:

The balanced equation for the reaction is given below:

Fe₂O₃ (s) + 2Al (s) → 2Fe (l) + Al₂O₃ (s)

Next, we shall determine the masses of Fe₂O₃ and Al that reacted and the masses of Fe and Al₂O₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Fe₂O₃ = (2×56) + (3×16)

= 112 + 48 = 160 g/mol

Mass of Fe₂O₃ from the balanced equation = 1 × 160 = 160 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

Molar mass of Fe = 56 g/mol

Mass of Fe from the balanced equation = 2 × 56 = 112 g

Molar mass of Al₂O₃ = (2×27) + (3×16)

= 54 + 48 = 102 g/mol

Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g

SUMMARY:

From the balanced equation above,

160 g of Fe₂O₃ reacted with 54 g of Al to produce 112 g of Fe and 102 g of Al₂O₃

A. Determination of the mass of iron(III) oxide, Fe₂O₃, needed to produce 76.73 g of iron, Fe.

From the balanced equation above,

160 g of Fe₂O₃ reacted to produce 112 g of Fe.

Therefore, Xg of Fe₂O₃ will react to produce 76.73 g of Fe i.e

Xg of Fe₂O₃ = (160 × 76.73)/112

Xg of Fe₂O₃ = 109.61 g

Thus, 109.61 g of Fe₂O₃ is needed to produce 76.73 g of Fe.

B. Determination of the mass of aluminum, Al, to produce 76.72 g of iron, Fe.

From the balanced equation above,

54 g of Al reacted to produce 112 g of Fe.

Therefore, Xg of Al will react to produce 76.72 g of Fe i.e

Xg of Al = (54 × 76.72)/112

Xg of Al = 36.99 g

Thus, 36.99 g of Al is needed to produce 76.72 g of Fe.

C. Determination of the maximum mass of aluminum oxide, Al₂O₃, produced along with 76.75 g Fe.

We'll begin by calculating the mass of Fe₂O₃ needed to produce 76.75 g of Fe. This is illustrated below:

From the balanced equation above,

160 g of Fe₂O₃ reacted to produce 112 g of Fe.

Therefore, Xg of Fe₂O₃ will react to produce 76.75 g of Fe i.e

Xg of Fe₂O₃ = (160 × 76.75)/112

Xg of Fe₂O₃ = 109.64 g

Thus, 109.64 g of Fe₂O₃ is needed to produce 76.75 g of Fe.

Finally, we shall determine the maximum mass of Al₂O₃ produced along with 76.75 g Fe. this can be obtained as follow:

From the balanced equation above,

160 g of Fe₂O₃ reacted 102 g of Al₂O₃.

Therefore, 109.64 g of Fe₂O₃ will react to produce = (109.64 × 102)/160 = 69.90 g of Al₂O₃.

Thus, the maximum mass of Al₂O₃ produced is 69.90 g