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(A) Calculate the moles of reactants:$Moles \ of \ Fe_3O_4 = \frac{2.32 	imes 10^3 	imes 10^3 \ g}{232 \ g \ mol^{-1}} = 10^4 \ mol$.$Moles \ of \ C

Vedclass · Accepted Answer

$420$

Vedclass · Answer

(A) Calculate the moles of reactants:$Moles \ of \ Fe_3O_4 = \frac{2.32 	imes 10^3 	imes 10^3 \ g}{232 \ g \ mol^{-1}} = 10^4 \ mol$.$Moles \ of \ CO = \frac{2.8 	imes 10^2 	imes 10^3 \ g}{28 \ g \ mol^{-1}} = 10^4 \ mol$.From the balanced equation $Fe_3O_4 + 4CO ightarrow 3Fe + 4CO_2$,the stoichiometric ratio is $1:4$.Since we have $10^4 \ mol$ of $Fe_3O_4$ and $10^4 \ mol$ of $CO$,$CO$ is the limiting reagent because it requires $4 	imes 10^4 \ mol$ of $CO$ to react completely with $10^4 \ mol$ of $Fe_3O_4$.$Moles \ of \ Fe \ produced = \frac{3}{4} 	imes (moles \ of \ CO) = \frac{3}{4} 	imes 10^4 = 7500 \ mol$.$Mass \ of \ Fe = 7500 \ mol 	imes 56 \ g \ mol^{-1} = 420000 \ g = 420 \ kg$.Therefore,$x = 420$.

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