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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2025 · 24 January · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

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 kgFe3O4and2.8×102 kgCO2.32 \times 10^3 \mathrm{~kg} \mathrm{Fe}_3 \mathrm{O}_4 and 2.8 \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 \left.\mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1}\right\}

Correct answer

0

Step-by-step solution

 moles of Fe3O4=2.32×103×103232=10000mol moles of CO=2.8×102×10328=10000molFe33O4+4CO⟶3Fe+4CO2104mol104mol\begin{aligned} & \text{ moles of } \mathrm{Fe}_3 \mathrm{O}_4=\frac{2.32 \times 10^3 \times 10^3}{232}=10000 \mathrm{ mol} \\ & \text{ moles of } \mathrm{CO}=\frac{2.8 \times 10^2 \times 10^3}{28}=10000 \mathrm{ mol} \\ & \mathrm{Fe}_3^3 \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=34×104×561000kg=420kg\mathrm{Fe}=\frac{3}{4} \times \frac{10^4 \times 56}{1000} \mathrm{ kg}=420 \mathrm{ kg}
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