Answer : The percent purity of
in the original sample is 87.94 %
Explanation :
The given balanced chemical reaction is:

First we have to calculate the mass of Fe.

Molar mass of Fe = 55.8 g/mole

Now we have to calculate the moles of 
From the balanced chemical reaction we conclude that,
As, 2 moles of Fe produced from 1 mole of 
So,
of Fe produced from
mole of 
Now we have to calculate the mass of 

Molar mass of
= 159.69 g/mole

Now we have to calculate the percent purity of
in the original sample.
Mass of original sample = 


Therefore, the percent purity of
in the original sample is 87.94 %