Question

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.

Answer 1

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($m_{Fe}$) = 565 × $\frac{1}{8}$ = 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 + $3O_{2}$ → $2Fe_{2} O_{3}$

Now, molecular weight of Iron = 55.84 g

So, moles of Iron undergoes the reaction is = $\frac{70.625}{55.84}$ = 1.26 mol

Molecular weight of $Fe_{2} O_{3}$ = 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 $Fe_{2} O_{3}$ = $\frac{159.68}{1.26}$ = 126.730 g

∴Mass of rust = 126.730 g

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