Final answer:
To answer how many moles of oxygen would be produced from 27.6 grams of aluminum, a specific chemical reaction involving aluminum is required. For water production, 13.8 moles of O2 are needed to produce 27.6 moles of H2O. In terms of aluminum, 27.6 grams correspond to approximately 1.024 moles of Al, dividing by its molar mass of 26.98 g/mol.
Step-by-step explanation:
The student appears to be asking about the stoichiometry of a chemical reaction involving aluminum and oxygen. However, the question mentioned seems to be a typo or mix-up, as there is no direct relation given between the mass of aluminum (Al) and the production of oxygen (O2). Instead, several references to exercises involving reactions to produce water (H2O) and aluminum oxide (Al2O3) are provided. To provide a precise answer, additional information about the chemical reaction involving aluminum would be needed.
Given the information provided, it's relevant to point out that the balanced chemical equation for the formation of water (2 H2 + O2 → 2 H2O) establishes that 1 mole of oxygen gas is required to produce 2 moles of water. Thus, to produce 27.6 moles of H2O, one would need 13.8 moles of O2. If the question intends to find out how many moles of oxygen would be produced from a reaction involving 27.6 grams of Al, we would need the specific chemical equation that describes this reaction.
For aluminum, 1 mole is equivalent to 26.98 grams. Therefore, the number of moles of aluminum in 27.6 grams would be calculated by dividing the mass by the molar mass (27.6 g ÷ 26.98 g/mol).