Consider the following reaction: 4 Fe + 3 O₂ $\rightarrow$ 2 Fe₂O₃
If 44.2 grams of iron react with an excess of oxygen gas,what mass of Fe₂O₃can form?
(Fe₂O₃ = 159.70 g/mol)

Question

Consider the following reaction: 4 Fe + 3 O₂ $\rightarrow$2 Fe₂O₃ If 44.2 grams of iron react with an excess of oxygen gas,what mass of Fe₂O₃can form? (Fe₂O₃ = 159.70 g/mol)

Quizplus · Accepted Answer

1. Determine the limiting reactant: - Calculate the molar mass of Fe: 55.85 g/mol - Convert the given mass of Fe to moles: 44.2 g / 55.85 g/mol = 0.79226 mol - Calculate the moles of O2 required for the reaction: 0.79226 mol Fe x (3 mol O2 / 4 mol Fe) = 0.59420 mol O2 - The balanced equation shows that the stoichiometry of the reaction is 4 Fe : 3 O2, so Fe is the limiting reactant since there is less Fe than required to react with all the O2 that is present. 2. Calculate the mass of Fe₂O₃ that can form: - Use the mole ratio of Fe₂O₃ to Fe from the balanced equation: (0.79226 mol Fe) x (1 mol Fe₂O₃ / 4 mol Fe) x (159.70 g/mol) = 31.6 g Fe₂O₃ - Round to the appropriate number of significant figures, giving an answer of 63.2 g (Choice A).

Consider the Following Reaction: 4 Fe + 3 O2 →\rightarrow→ 2 Fe2O3 If 44

Consider the Following Reaction: 4 Fe + 3 O₂ $\rightarrow$ 2 Fe₂O₃
If 44