Consider the following reaction occurring in the blast furnace :\ $\mathrm{Fe}_{3} \mathrm{O}_{4(s)

Question

Consider the following reaction occurring in the blast furnace :\ $\mathrm{Fe}_{3} \mathrm{O}_{4(s)}+4 \mathrm{CO}_{(g)} ightarrow 3 \mathrm{Fe}_{(l)}+4 \mathrm{CO}_{2(g)}$\ ' $x$ ' kg of iron is produced when $2.32 	imes 10^{3} \mathrm{~kg} \mathrm{Fe}_{3} \mathrm{O}_{4}$ and $2.8 	imes 10^{2} \mathrm{~kg} \mathrm{CO}$ are brought together in the furnace. The value of ' $x$ ' is $\_\_\_\_$ . (nearest integer)\ [Given : molar mass of $\mathrm{Fe}_{3} \mathrm{O}_{4}=232 \mathrm{~g} \mathrm{~mol}^{-1}$ molar mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$\ molar mass of $\mathrm{Fe}=56 \mathrm{~g} \mathrm{~mol}^{-1}$ ]\

Professor PC Thomas Classes · Answer

$$ \begin{aligned}  	ext {  (420) : } \mathrm{Fe}_{3} \mathrm{O}_{4(s)}+4 \mathrm{CO}_{(g)} \longrightarrow \  2.32 	imes 10^{3} \mathrm{~kg} \mathrm{Fe}_{(l)}+4 \mathrm{CO}_{2(g)}= \  \frac{2.32 	imes 10^{3} 	imes 10^{3}}{232}=10^{4} \mathrm{~mole} \  2.8 	imes 10^{2} \mathrm{~kg} \mathrm{CO}= \ & \frac{2.8 	imes 10^{2} 	imes 10^{3}}{28}=10^{4} \mathrm{~mole} \end{aligned} $$ Here CO will be limiting reagent as for $10^{4}$ moles of $\mathrm{Fe}_{3} \mathrm{O}_{4} 4 	imes 10^{4}$ moles of CO are required.\ Amount of Fe produced $$ \begin{aligned} & =\frac{3}{4} 	imes 10^{4} 	ext { moles } \ & =\frac{3}{4} 	imes 10^{4} 	imes 56 \mathrm{~g} \ & =\frac{3}{4} 	imes 560 \mathrm{~kg}=420 \mathrm{~kg} \end{aligned} $$