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When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? $4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)$?

Md Ariful Islam

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Md Ariful Islam

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We have the balanced equation:

$4Fe(s)+3O_2(g)->2Fe_2O_3(s)$

And so, $4$ moles of iron react with $3$ moles of oxygen, and therefore, $12.9 \ "mol"$ of iron would react with:

$12.9color(red)cancelcolor(black)("mol" \ Fe)*(3 \ "mol" \ O_2)/(4color(red)cancelcolor(black)("mol" \ Fe))=9.675 \ "mol" \ O_2$

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