ifa piece of iron gains 10 percent of its weight due to partial rusting into fe2o3 then what is the percentage of total iron rusted

 

Asked by tiwariakanksha0103 | 24th Jul, 2019, 07:46: PM

Expert Answer:

The chemical reaction for rusting of iron is as follows:

4Fe + 3O2 → 2Fe2O3

Initial mass of iron =100 gm

Mass of Fe2O3 = 110 gm

Mass of oxide = 10 gm

Now, from equation,

To form 2 moles of rust it requires 3 moles of O2 requires 4 moles of Fe

10/32 moles will use

equals space 4 over 3 cross times 10 over 32 space moles space of space Fe

equals 40 over 96

equals space 0.4166 space mole space of space Fe

 

Mass of Fe reacted = 0.4166×56

                            =23.33%

Answered by Varsha | 25th Jul, 2019, 10:10: AM

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