Production of Fe in blast furnace: Fe3O4(s)+4CO(g)-3Fe(l)+4CO2(g). When 4.640\,kg of Fe3O4 and 2.520\,kg of CO are…

🧠 Compare moles, not kilograms The reaction needs $1$ mole of $\ce{Fe3O4}$ for every $4$ moles of $\ce{CO}$. Convert both reactants to moles, find which one runs out first, then read the iron yield off the balanced equation.

🗺️ Step by step Molar mass of $\ce{Fe3O4} = 3(56) + 4(16) = 232\,\text{g/mol}$. Moles of $\ce{Fe3O4} = 4640 / 232 = 20\,\text{mol}$. Moles of $\ce{CO} = 2520 / 28 = 90\,\text{mol}$.

For $20\,\text{mol}$ of oxide you need $4 \times 20 = 80\,\text{mol}$ of CO. You have $90$, so $\ce{CO}$ is in excess. $\ce{Fe3O4}$ is the limiting reagent.

Each $\ce{Fe3O4}$ gives $3$ atoms of Fe: $\text{Moles Fe} = 20 \times 3 = 60\,\text{mol}.$ $\text{Mass Fe} = 60 \times 56 = 3360\,\text{g}.$

⚡ Fast check $4640 = 20 \times 232$ exactly. So you start with $20$ neat moles of oxide. Triple it for Fe atoms, multiply by $56$, done.

$\boxed{\text{Answer: (c)}}$