Limiting Reagent In Multi-step Context

Identifying the limiting reagent is the critical first step in any yield calculation. Moles of (\text{Fe}_2\text{O}_3 = 160/160 = 1.0) mol. Moles of CO (= 84/28 = 3.0) mol. From the balanced equation, 1 mol (\text{Fe}_2\text{O}_3) requires 3 mol CO. Available CO (= 3.0) mol, which is exactly the stoichiometric requirement for 1 mol (\text{Fe}_2\text{O}_3). Both reactants are consumed in exact stoichiometric amounts, so neither is in excess. From 1 mol (\text{Fe}_2\text{O}_3), the equation gives 2 mol Fe. Mass of Fe (= 2 \times 56 = 112) g. Option 56 g corresponds to only 1 mol Fe, which would arise from assuming 1:1 stoichiometry instead of 1:2 for (\text{Fe}_2\text{O}_3):(\text{Fe}). Option 168 g would require 3 mol Fe, but the equation yields only 2 mol per mol of (\text{Fe}_2\text{O}_3). Option 224 g would require 4 mol Fe, which would need 2 mol (\text{Fe}_2\text{O}_3), but only 1 mol is available. This is an NCERT-pattern stoichiometry question embedded in an industrially significant context (blast furnace iron extraction) common in JEE. Plausibility check: 112 g of Fe from 160 g of (\text{Fe}_2\text{O}_3) represents (2 \times 56/160 = 70%) yield by mass, consistent with Fe being (2 \times 56/160 \times 100 = 70%) of (\text{Fe}_2\text{O}_3).