IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

punnima

a mixture of FeO and Fe3O4 is heated in d presence of O2 and it resulted in a 5%inc in mass
wat is d original composition???

EDITED the Qn.

punnima

sum plizz gimme an ans

sandeepramesh

first write the oxidative rns here

punnima

i didnn get u......
i didnn even understand d prob

eistien

see in the above question the increase in mass is due to oxidation of FeO to Fe2O3 consider 100 g of the originale composition with x gms FeO and 100-x gms of Fe2O3 and then now finally the compisition weighs 105 gms and x gms of FeO has reacted to give Fe2O3 and thus u can solve for x by equating in the second equation in using that finall composition weighs 105 gms do you want me to calculate and tell the answer!!! this is enough no???

punnima

plizz write d reacns d d calc...it will b of gr8 help

eistien

dei rohit this is our santa sir's mole concept sheet 4th sum da refer there if u have a doubt!!!

eistien

nways here it is see

4FeO + 3O2-> 2Fe2O3

x (3x/4) (x/2)

so see now total amount left after the reaction is (x/2 + 100-x) of Fe2O3 so this is equal to 105g

so (100-x/2)(mass of Fe2O3)=105 gms

solve for it and you get the answer!!!

punnima

i edited d qn ...it was sposed 2 b Fe3O4

eistien

that makes it much more easier

see here for that sum

4FeO + O2-> 2Fe2O3

4Fe3O4+O2-> 6Fe2O3

let FeO weigh x gms

Fe3O4 weigh 100-x gms

2(56+16) of FeO corresponds to 2*56+3*16 of Fe2O3(from 1st equation)

2(3*56+4*16) of Fe3O4 corresponds to 2*56+3*16 of Fe2O3(from 2ndequation)

so equating the corresponding ones

we get

(160x/144)+(480/464)(100-x) gives 105 grams

solve for x!!!

eistien

hey the composition is 21.0526% of FeO and 78.9474% of Fe3O4

sandeepramesh

that is answer nv :D :D :D

eistien

thanks for your encouraging comments ROHIT!!!

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