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Answer:
674.26 g of AlI₃
Explanation:
We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:
Percentage yield of Al = 67.8%
Actual yield of Al = 30.25 g
Theoretical yield of Al =?
Percentage yield = Actual yield /Theoretical yield × 100/
67.8% = 30.25 / Theoretical yield
67.8 / 100 = 30.25 / Theoretical yield
0.678 = 30.25 / Theoretical yield
Cross multiply
0.678 × Theoretical yield = 30.25
Divide both side by 0.678
Theoretical yield = 30.25 / 0.678
Theoretical yield of Al = 44.62 g
Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:
AlI₃(s) + 3K(s) → 3KI(s) + Al(s)
Molar mass of AlI₃ = 27 + (3×127)
= 27 + 381 = 408 g/mol
Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g
Molar mass of Al = 27 g/mol
Mass of Al from the balanced equation = 1 × 27 = 27 g
Summary:
From the balanced equation above,
408 g of AlI₃ reacted to produce 27 g of Al.
Finally, we shall determine the mass of
AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:
From the balanced equation above,
408 g of AlI₃ reacted to produce 27 g of Al.
Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e
Xg of AlI₃ = (408 × 44.62)/27
Xg of AlI₃ = 674.26 g
Thus, 674.26 g of AlI₃ is needed for the reaction.