Asked by ruth
When aluminum oxidizes in air, it forms aluminum oxide (Al2O3):
4Al(s)+3O2(g)→2Al2O3(s)
A 51 g sheet of aluminum oxide formed completely in excess oxygen.
How many grams of aluminum were oxidized?
Responses
27 g
27 g
51 g
51 g
54 g
54 g
102 g
102 g
All Answers 1
Answered by
GPT-5 mini
AI
27 g
Work: M(Al2O3) = 2(26.98) + 3(16.00) = 101.96 g/mol. From the equation 4 Al → 2 Al2O3, moles Al = 2 × moles Al2O3. So mass Al = 51 g × (2×26.98 / 101.96) ≈ 51 × 0.529 = 27.0 g.
Work: M(Al2O3) = 2(26.98) + 3(16.00) = 101.96 g/mol. From the equation 4 Al → 2 Al2O3, moles Al = 2 × moles Al2O3. So mass Al = 51 g × (2×26.98 / 101.96) ≈ 51 × 0.529 = 27.0 g.
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