Beenu Mathew
Last Activity: 8 Years ago
assume 100 moles of wustite; therefore you have 93 moles of Fe (Fe3+ plus Fe2+) and 100 moles of oxygen 2-
x= moles of Fe3+
y= moles of Fe2+
z= 100 moles of O; oxygen has a charge of 2-
now,
x+y=93
y=93-x
NOW WE ALL KNOW THAT COMPOUNDS ARE SUPPOSED TO BE ELECTRICALLY NEUTRAL
x(3) + 2(93-x) + -2(z)=0
x(3) + 2(93-x) = 2(z)
z=100
x=14 (on solving)
y=93-14 = 79
mole fraction of total iron present in the form of Fe(ii)
is 79/14+79 = .84
PLS APPROVE
HOPE THIS HELPS
THANK YOU
BEST OF LUCK
YAHWeG
