Answer:
A. 109.61 g of Fe₂O₃
B. 36.99 g of Al
C. Maximum mass of Al₂O₃ produced is 69.90 g
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
