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Production of iron in blast furnace follows the following equation

$\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+4 \mathrm{CO}(\mathrm{g}) \rightarrow 3 \mathrm{Fe}(\mathrm{l})+4 \mathrm{CO}_{2}(\mathrm{~g})$

when $4.640 \mathrm{~kg} \text { of } \mathrm{Fe}_{3} \mathrm{O}_{4} \text { and } 2.520 \mathrm{~kg} \text { of } \mathrm{CO}$ are allowed to react then the amound of iron (in g) produced is:

$\text { [Given: } \quad \text { Molar Atomic mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right): \mathrm{Fe}=56$

$\text { Molar Atomic mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right): \mathrm{O}=16$

$\text { Molar Atomic mass } \left.\left(\mathrm{g} \mathrm{mol}^{-1}\right): \mathrm{C}=12\right]$
Option: 1
$1400$

Option: 2
$2200$

Option: 3
$3360$

Option: 4
$4200$

Answers (1)

best_answer

$\mathrm{moles\: of\: \mathrm{Fe}_{3} \mathrm{O}_{4}=\frac{4640}{232}=20}\\$

$\mathrm{moles\: of\: C O=\frac{2520}{28}=90}\\$

$\mathrm{\therefore Limiting \: reagent \: is \: \mathrm{Fe}_{3} \mathrm{O}_{4}}$

$\mathrm{\mathrm{Fe}_{3} \mathrm{O}_{4}+4 \mathrm{CO} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_{2}}$

From reaction stoichiometry,

1 mole $\mathrm{\mathrm{Fe}_{3} O_{4}}$ produces 3 moles of $\mathrm{\mathrm{Fe}}$

$\mathrm{\therefore\text{ moles of Fe produced}=60}$

$\therefore\mathrm{\text{mass of Fe produced}=3360}$

Hence the correct answer is option 3

Posted by

Divya Prakash Singh

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