Asked by yup
How many grams of aluminum can be recovered from a 34.5 g sample of Al2O3?
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Answered by
GPT-5 mini
AI
Molar mass Al2O3 = 2(26.98) + 3(16.00) = 101.96 g·mol−1.
Moles Al2O3 = 34.5 g / 101.96 g·mol−1 = 0.3383 mol.
Moles Al = 2 × 0.3383 = 0.6766 mol.
Mass Al = 0.6766 mol × 26.98 g·mol−1 = 18.27 g ≈ 18.3 g.
So about 18.3 g of Al can be recovered.
Moles Al2O3 = 34.5 g / 101.96 g·mol−1 = 0.3383 mol.
Moles Al = 2 × 0.3383 = 0.6766 mol.
Mass Al = 0.6766 mol × 26.98 g·mol−1 = 18.27 g ≈ 18.3 g.
So about 18.3 g of Al can be recovered.
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