If aluminum loses 3.01times 1024 electrons in the reaction with each of oxygen and fluorine gases, the ratio of the mass of aluminum fluoride produced to -r What is the mass of aluminum oxide produced?

To solve this problem, we need to use the balanced chemical equations for the reactions of aluminum with oxygen and fluorine, and then calculate the mass of aluminum oxide produced.<br /><br />Given information:<br />- Aluminum loses $3.01\times 10^{24}$ electrons in the reaction with oxygen and fluorine gases.<br /><br />Step 1: Write the balanced chemical equations for the reactions.<br />Aluminum reacts with oxygen gas to form aluminum oxide:<br />4Al + 3O2 → 2Al2O3<br /><br />Aluminum reacts with fluorine gas to form aluminum fluoride:<br />2Al + 3F2 → 2AlF3<br /><br />Step 2: Calculate the number of moles of aluminum.<br />The number of moles of aluminum can be calculated using the formula:<br />Moles of aluminum = Number of electrons / (Avogadro's number × Charge of aluminum ion)<br />Moles of aluminum = $3.01\times 10^{24}$ / (6.022 × 10^23 × 3)<br />Moles of aluminum = 1.66 mol<br /><br />Step 3: Calculate the mass of aluminum oxide produced.<br />The balanced equation shows that 4 moles of aluminum produce 2 moles of aluminum oxide.<br />Moles of aluminum oxide = 1.66 mol × (2/4) = 0.83 mol<br />Mass of aluminum oxide = Moles of aluminum oxide × Molar mass of aluminum oxide<br />Mass of aluminum oxide = 0.83 mol × (26.98 g/mol × 2) = 44.8 g<br /><br />Therefore, the mass of aluminum oxide produced is 44.8 g.