The correct answer to the question is 65.4 g
The balanced equation for the reaction is given below
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Next, we shall determine the mass of Al₂O₃ that reacted and the mass of AlCl₃ produced from the balanced.
Molar mass of Al₂O₃ = (27×2) + (3×16)
= 54 + 48
= 102 g/mol
Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g
Molar mass of AlCl₃ = 27 + (35.5×3)
= 27 + 106.5
= 133.5 g/mol
Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g
Thus,
From the balanced equation above,
102 g of Al₂O₃ reacted to produce 267 g of AlCl₃.
Finally, we shall determine the mass of AlCl₃ produced by the reaction of 35 g of Al₂O₃. This is illustrated below:
From the balanced equation above,
102 g of Al₂O₃ reacted to produce 267 g of AlCl₃.
Therefore,
25 g of Al₂O₃ will react to produce = [tex]\frac{25 * 267}{102}[/tex] = 65.4 g of AlCl₃.
Thus, 65.4 g of AlCl₃ were obtained from the reaction.
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