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Answer :
Final answer:
The maximum mass of iron(III) oxide that can be formed is 71.86 grams.
Explanation:
In this reaction, we have 38.1 grams of iron and 14.4 grams of oxygen gas. To determine the maximum mass of iron(III) oxide that can be formed, we need to identify the limiting reactant.
To do this, we can calculate the number of moles of each reactant using their molar masses. The molar mass of iron is 55.85 g/mol, and the molar mass of oxygen gas is 32.00 g/mol.
For iron:
Number of moles = mass / molar mass = 38.1 g / 55.85 g/mol = 0.682 mol
For oxygen gas:
Number of moles = mass / molar mass = 14.4 g / 32.00 g/mol = 0.450 mol
Next, we need to determine the mole ratio between iron and iron(III) oxide in the balanced chemical equation. From the equation, we can see that the ratio is 1:1.
Since the mole ratio is 1:1, the limiting reactant is the reactant that has the smaller number of moles. In this case, oxygen gas has 0.450 moles, which is smaller than the 0.682 moles of iron.
Now that we know the limiting reactant is oxygen gas, we can use stoichiometry to calculate the maximum mass of iron(III) oxide that can be formed.
From the balanced chemical equation, we can see that the molar mass of iron(III) oxide is 159.69 g/mol.
Number of moles of iron(III) oxide = number of moles of oxygen gas = 0.450 mol
Mass of iron(III) oxide = number of moles * molar mass = 0.450 mol * 159.69 g/mol = 71.86 g
Therefore, the maximum mass of iron(III) oxide that can be formed is 71.86 grams.
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