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(C) $1$. Calculate the molar mass of $Fe_{3}O_{4} = (3 	imes 56) + (4 	imes 16) = 168 + 64 = 232 \ g \ mol^{-1}$.$2$. Calculate the moles of $Fe_{3}

Vedclass · Accepted Answer

$3360$

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(C) $1$. Calculate the molar mass of $Fe_{3}O_{4} = (3 	imes 56) + (4 	imes 16) = 168 + 64 = 232 \ g \ mol^{-1}$.$2$. Calculate the moles of $Fe_{3}O_{4} = \frac{4.640 	imes 10^{3} \ g}{232 \ g \ mol^{-1}} = 20 \ mol$.$3$. Calculate the molar mass of $CO = 12 + 16 = 28 \ g \ mol^{-1}$.$4$. Calculate the moles of $CO = \frac{2.520 	imes 10^{3} \ g}{28 \ g \ mol^{-1}} = 90 \ mol$.$5$. According to the stoichiometry,$1 \ mol$ of $Fe_{3}O_{4}$ requires $4 \ mol$ of $CO$. Thus,$20 \ mol$ of $Fe_{3}O_{4}$ requires $20 	imes 4 = 80 \ mol$ of $CO$.$6$. Since we have $90 \ mol$ of $CO$ (which is more than $80 \ mol$),$Fe_{3}O_{4}$ is the limiting reagent.$7$. From the equation,$1 \ mol$ of $Fe_{3}O_{4}$ produces $3 \ mol$ of $Fe$. Therefore,$20 \ mol$ of $Fe_{3}O_{4}$ produces $20 	imes 3 = 60 \ mol$ of $Fe$.$8$. Mass of $Fe$ produced $= 60 \ mol 	imes 56 \ g \ mol^{-1} = 3360 \ g$.

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Assertion : The amount of the products formed in a reaction depends upon the limiting reagent.
Reason : Limiting reactant reacts completely in the reaction therefore limits the amount of product.

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