Question

iron (III) oxide is formed according to the following balanced chemical equation:
\[{\text{4F}}{{\text{e}}_{{\text{(g)}}}}\,{\text{ + }}\,{\text{3}}{{\text{O}}_{{\text{2(g)}}}}\, \to
2\,{\text{F}}{{\text{e}}_2}{{\text{O}}_{{\text{3(g)}}}}\]
The reaction of $8.0$ moles of solid iron with $9.0$moles of oxygen gas produces:

A. \[4.0\] moles of \[{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\] and \[3.0\] moles of excess
\[{{\text{O}}_2}\].
B. \[4.0\] moles of \[{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\] and \[6.0\] moles of excess
\[{{\text{O}}_2}\].
C. \[6.0\] moles of \[{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\] and \[2.0\] moles of excess
\[{{\text{O}}_2}\].
D. \[6.0\] moles of \[{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\] and \[3.0\]moles of excess
\[{{\text{O}}_2}\].
E. \[6.0\] moles of \[{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\] and \[4.0\]moles of excess
\[{{\text{O}}_2}\].

Answer 1

Hint: By comparing the mole ratio, the amount of product is determined. The reactants react in some ratio so one species that is used completely known as the limiting agent and the one which is in excess is left after the reaction.

Step by step answer: Determine the amount of product by comparing the mole of solid iron and product iron (III) oxide as follows:
Four moles of solid iron gives two moles of iron (III) oxide so, determine the amount of iron (III) oxide produced by eight moles of the solid iron as follows:
 \[\, = \dfrac{{2\,{\text{mol}}\,\,{\text{F}}{{\text{e}}_2}{{\text{O}}_{\text{3}}}\,}}{{{\text{4mol}}\,\,{\text{Fe}}}} \times \,{\text{8 mol}}\,\,{\text{Fe}}\]
\[\,{\text{Mole}}\,\,{\text{of}}\,\,{\text{F}}{{\text{e}}_2}{{\text{O}}_3} = 4\,{\text{mol}}\,\,{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\]
So, eight moles of solid iron gives four moles of iron (III) oxide.
Determine the amount of oxygen left as follows:
Four moles of solid iron react with three moles of oxygen so, determine the amount of oxygen reacts with eight moles of the solid iron as follows:
\[\,{\text{Mole}}\,\,{\text{of}}\,{{\text{O}}_2} = \dfrac{{3\,\,{\text{mol}}\,{{\text{O}}_2}\,}}{{4\,{\text{mol}}\,\,{\text{Fe}}}} \times \,{\text{8mol}}\,\,{\text{Fe}}\]
\[\,{\text{Mole}}\,\,{\text{of}}\,{{\text{O}}_2} = \,6\,{\text{Mol}}\,{{\text{O}}_2}\]
So, six moles of oxygen will react with eight moles of solid iron.
The given amount of oxygen is nine moles out of which six moles react with solid iron so, the amount of oxygen left is.
\[\,{\text{Mole}}\,\,{\text{of}}\,{{\text{O}}_2}\,{\text{left}} = \,\,9\,{\text{Mol}}\,{{\text{O}}_2}\, - 6\,{\text{Mol}}\,{{\text{O}}_2}\]
\[\,{\text{Mole}}\,\,{\text{of}}\,{{\text{O}}_2}\,{\text{left}} = \,\,3\,{\text{Mol}}\,{{\text{O}}_2}\]
So, the three moles oxygen is in excess.

Therefore, option (A) \[4.0\]moles of \[{\text{F}}{{\text{e}}_2}{{\text{O}}_3}\] and \[3.0\]moles of excess \[{{\text{O}}_2}\], is correct.

Note: The reactants react in some ratio so, to determine the amount of one reactant from other reactants the balanced reaction is required. Limiting reagent is the species that is present in a low amount in the reaction decides the amount of product and the amount of another reactant that will be used in the reaction.