Top Posters
Since Sunday
g
3
3
2
J
2
p
2
m
2
h
2
s
2
r
2
d
2
l
2
a
2
Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
New Topic  
Dianajupiter1 Dianajupiter1
wrote...
Go to Answer
Posts: 120
Rep: 0 0
3 years ago
You heat 3.886 g of a mixture of Fe3O4 and FeO to form 4.095 g Fe2O3. The mass percent of FeO originally in the mixture was


▸ 94.9%.

▸ 26.1%.

▸ 73.9%.

▸ 18.0%.

▸ None of these are correct.
Textbook 
Chemistry: An Atoms First Approach

Chemistry: An Atoms First Approach


Edition: 3rd
Authors:
Read 208 times
2 Replies
Replies
Answer verified by a subject expert
college98college98
wrote...
#1
Posts: 112
Rep: 2 0
3 years ago
Sign in or Sign up in seconds to unlock everything for free
More solutions for this book are available here
1

Related Topics

wrote...
#2
A year ago
Fe2O3 moles = mass / Molar mass of Fe2O3

= 4.095 g / 159.7 g/mol

= 0.02564

Let FeO moles = m , Fe3O4 moles = 0.02564-m

Mass of FeO = moles x molar mass of FeO

= m x 71.844

mass of Fe3O4 = (0.02564 -m) x 231.533 g/mol

= 5.9365 - 231.533m

total mass of FeO + Fe3O4 = 3.886

71.844 m + 5.9365 - 231.533 m = 3.886

m = 0.01284

FeO mass = 71.844 g/mol x 0.01284 mol = 0.9225g

mass % of FeO = ( 100 x FeO mass/ sample mass)

= ( 100 x 0.9225 g / 3.886 g)

= 23.7 %
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  1121 People Browsing
 137 Signed Up Today
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 256
  
 1397
  
 82