Final answer:
Approximately 39.94 grams of iron oxide will be produced when 28 grams of iron reacts with 16 grams of oxygen.
Step-by-step explanation:
To determine how much iron oxide will be produced when 28 grams of iron reacts with 16 grams of oxygen, we need to use the balanced chemical equation for the reaction. The balanced equation is:
4 Fe + 3 O2 → 2 Fe2O3
From the balanced equation, we can see that the mole ratio between iron and iron oxide is 4:2. Therefore, for every 4 moles of iron, 2 moles of iron oxide will be produced. We can calculate the moles of iron and moles of oxygen using their respective masses:
Moles of iron = (Mass of iron) / (Molar mass of iron)
Moles of oxygen = (Mass of oxygen) / (Molar mass of oxygen)
Then, we can use the mole ratio to determine the moles of iron oxide produced:
Moles of iron oxide = (Moles of iron) × (Mole ratio)
Finally, we can convert the moles of iron oxide produced into grams:
Mass of iron oxide = (Moles of iron oxide) × (Molar mass of iron oxide)
Plugging in the values:
Moles of iron = 28 g / 55.85 g/mol = 0.501 mol
Moles of oxygen = 16 g / 16.00 g/mol = 1 mol
Moles of iron oxide = 0.501 mol × (2/4) = 0.2505 mol
Mass of iron oxide = 0.2505 mol × 159.69 g/mol = 39.94 g
Therefore, approximately 39.94 grams of iron oxide will be produced.