Calculate the weight of FeO produced from 6.7 | vedclass

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(A) The balanced chemical equation is: $2VO + 3Fe_2O_3 \rightarrow 6FeO + V_2O_5$ Calculate the moles of reactants: $Moles \ of \ VO = \frac{6.7 \ g}{

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$4.32$

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(A) The balanced chemical equation is: $2VO + 3Fe_2O_3 ightarrow 6FeO + V_2O_5$ Calculate the moles of reactants: $Moles \ of \ VO = \frac{6.7 \ g}{67 \ g/mol} = 0.1 \ mol$ $Moles \ of \ Fe_2O_3 = \frac{4.8 \ g}{160 \ g/mol} = 0.03 \ mol$ Determine the limiting reagent: For $VO$: $\frac{0.1}{2} = 0.05$ For $Fe_2O_3$: $\frac{0.03}{3} = 0.01$ Since $0.01 Calculate moles of $FeO$ produced: From the stoichiometry,$3 \ mol \ Fe_2O_3$ produces $6 \ mol \ FeO$. $Moles \ of \ FeO = 2 	imes Moles \ of \ Fe_2O_3 = 2 	imes 0.03 = 0.06 \ mol$ Calculate weight of $FeO$: $Weight = Moles 	imes Molar \ Mass = 0.06 \ mol 	imes 72 \ g/mol = 4.32 \ g$

Calculate the weight of $FeO$ produced from $6.7 \ g$ of $VO$ and $4.8 \ g$ of $Fe_2O_3$ using the reaction: $2VO + 3Fe_2O_3 \rightarrow 6FeO + V_2O_5$ $(V = 51, Fe = 56)$ (in $g$)

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