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An iron bar weighed 565 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe2O3). Calculate the final mass of the iron bar and rust.

Respuesta :

Answer:

Final mass of Iron bar will be 621.105 g.

Mass of rust = 126.730 g

Explanation:

Exactly one- eighth of the Iron turns to rust. That means, the mass of Iron  that under go the process of rusting([tex]m_{Fe}[/tex]) = 565 × [tex]\frac{1}{8}[/tex] = 70.625 g

So, the mass of pure Iron after rusting will be = 565 - 70.625 = 494.375 g

Now the reaction of rusting as follows,

   2Fe + [tex]3O_{2}[/tex] → [tex]2Fe_{2} O_{3}[/tex]

Now, molecular weight of Iron = 55.84 g

So, moles of Iron undergoes the reaction is = [tex]\frac{70.625}{55.84}[/tex] = 1.26 mol

Molecular weight of [tex]Fe_{2} O_{3}[/tex] = 159.68 g

Here 2 moles of Iron undergoes the reaction to produce 2 moles of Iron oxide.

Hence, 1.26 mol of Iron will produce 1.26 mole of Iron oxide.

Mass of [tex]Fe_{2} O_{3}[/tex] = [tex]\frac{159.68}{1.26}[/tex] = 126.730 g

∴Mass of rust = 126.730 g

∴ Final mass of Iron bar = ( 494.375 + 126.730) = 621.105 g

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