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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,$x = 420$.

Similar Questions

In the reaction $4NH_{3(g)} + 5O_{2(g)} \rightarrow 4NO_{(g)} + 6H_2O_{(g)}$,when $1 \ mol$ of ammonia and $1 \ mol$ of $O_2$ react to completion:

Identify the values of $A$, $B$, $C$, $D$ and $E$ for the following unbalanced reaction: $aKNO_3 + 5C_{12}H_{22}O_{11} \rightarrow bN_2 + cCO_2 + dH_2O + eK_2CO_3$

$50.0 \ kg$ of $N_{2(g)}$ and $10.0 \ kg$ of $H_{2(g)}$ are mixed to produce $NH_{3(g)}$. Calculate the amount of $NH_{3(g)}$ formed. Identify the limiting reagent in the production of $NH_{3(g)}$ in this situation.

$3O_2 + 2N_2 \to 2N_2O_3$
$9 \ mol$ of $O_2$ and $14 \ mol$ of $N_2$ are allowed to react. When $3 \ mol$ of $O_2$ remains unreacted,how many moles of $N_2O_3$ would have been produced?

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