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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 July 2019, 7:46 PM
Answered by 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 Expert 25th July 2019, 10:10 AM
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