Verified Solution • 2025–26 Session moderate

QJEEM25Chf48b: JEE Main 2025 Chemistry - Class 11 JEEM25Chf48b Solution: Consider the following reaction occurring in the blast furnace: …

Consider the following reaction occurring in the blast furnace:

Fe3O4( s)+4CO(g)→3Fe(l)+4CO2( g)\mathrm{Fe}_3 \mathrm{O}_{4(\mathrm{~s})}+4 \mathrm{CO}_{(\mathrm{g})} \rightarrow 3 \mathrm{Fe}_{(\mathrm{l})}+4 \mathrm{CO}_{2(\mathrm{~g})}

' xx ' kg of iron is produced when 2.32×103 kgFe3O42.32 \times 10^3 \mathrm{~kg} \mathrm{Fe}_3 \mathrm{O}_4 and 2.8×102 kgCO2.8 \times 10^2 \mathrm{~kg} \mathrm{CO} are brought together in the furnace. The value of ' xx ' is _________ . (nearest integer)

{Given: molar mass of Fe3O4=232 g mol−1\mathrm{Fe}_3 \mathrm{O}_4=232 \mathrm{~g} \mathrm{~mol}^{-1}

molar mass of CO=28 g mol−1\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}

molar mass of Fe=56 g mol−1\mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1} }

Step-by-Step Verified Solution

moles of Fe3O4=2.32×103×103232=10000 mol\mathrm{Fe}_3 \mathrm{O}_4=\frac{2.32 \times 10^3 \times 10^3}{232}=10000 \mathrm{~mol}

moles of CO=2.8×102×10328=10000 mol\mathrm{CO}=\frac{2.8 \times 10^2 \times 10^3}{28}=10000 \mathrm{~mol}

amp;Fe3O4+4CO⟶3Fe+4CO2amp;104 mol104 mol\begin{aligned}& \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_2 \\& 10^4 \mathrm{~mol} \quad 10^4 \mathrm{~mol}\end{aligned}

CO is L.R.

mole of Fe=34×104\mathrm{Fe}=\frac{3}{4} \times 10^4

mass of Fe=34×104×561000 kg=420 kg\mathrm{Fe}=\frac{3}{4} \times \frac{10^4 \times 56}{1000} \mathrm{~kg}=420 \mathrm{~kg}

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