a. To determine the limiting reactant, we need to compare the amount of each reactant to their stoichiometric coefficients in the balanced equation.
Given:
Mass of iron (Fe) = 2.8 g
Molar mass of iron (Fe) = 56 g/mol
Mass of oxygen (O₂) = 1.28 g
Molar mass of oxygen (O₂) = 32 g/mol
Using the molar masses, we can convert the masses to moles:
Moles of iron (Fe) = 2.8 g / 56 g/mol = 0.05 mol
Moles of oxygen (O₂) = 1.28 g / 32 g/mol = 0.04 mol
From the balanced equation, we can see that the stoichiometric ratio between iron and oxygen is 4:3. This means that for every 4 moles of iron, we need 3 moles of oxygen.
Comparing the moles of iron and oxygen, we find that the ratio of moles is 0.05:0.04. Since the ratio is not 4:3, the limiting reactant is the one that produces fewer moles. In this case, oxygen (O₂) is the limiting reactant.
b. To find the mass of Fe₂O₃ formed, we can use the stoichiometry of the balanced equation.
From the balanced equation:
4 moles of Fe → 2 moles of Fe₂O₃
Since we have determined that oxygen is the limiting reactant, we can use its moles to calculate the moles of Fe₂O₃ formed.
Moles of Fe₂O₃ = (0.04 mol O₂) * (2 mol Fe₂O₃ / 3 mol O₂) = 0.0267 mol Fe₂O₃
Now, we can calculate the mass of Fe₂O₃ using its molar mass:
Mass of Fe₂O₃ = 0.0267 mol * 160 g/mol = 4.27 g
Therefore, approximately 4.27 grams of Fe₂O₃ will be formed in the reaction.
c. Since oxygen is the limiting reactant, it will be completely consumed in the reaction. Therefore, there will be no excess oxygen remaining.
To calculate the mass of the excess iron (Fe) that remains after the reaction, we can determine how much iron reacted by subtracting the moles of Fe₂O₃ formed from the moles of iron initially present.
Moles of iron reacted = (0.05 mol Fe) - (0.0267 mol Fe₂O₃ / 2 mol Fe) = 0.0366 mol Fe
Now, we can calculate the mass of the excess iron:
Mass of excess Fe = 0.0366 mol * 56 g/mol = 2.05 g
Therefore, approximately 2.05 grams of excess iron will remain after the reaction stops.