menu
QAmmunity.org
Login
Register
My account
Edit my Profile
Private messages
My favorites
Register
Ask a Question
Questions
Unanswered
Tags
Categories
Ask a Question
A chemist heats 100.0 g of FeSO4 x 7H2O in a crucible to drive off the water. If all the water is driven off, what is the mass of the remaining salt?
asked
Aug 16, 2018
113k
views
4
votes
A chemist heats 100.0 g of FeSO4 x 7H2O in a crucible to drive off the water. If all the water is driven off, what is the mass of the remaining salt?
Chemistry
high-school
Alexandre Passos
asked
by
Alexandre Passos
6.9k
points
answer
comment
share this
share
0 Comments
Please
log in
or
register
to add a comment.
Please
log in
or
register
to answer this question.
2
Answers
2
votes
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
Justin Talbott
answered
Aug 18, 2018
by
Justin Talbott
7.0k
points
ask related question
comment
share this
0 Comments
Please
log in
or
register
to add a comment.
5
votes
The correct answer is 54.6g
Sergey Slepov
answered
Aug 22, 2018
by
Sergey Slepov
6.0k
points
ask related question
comment
share this
0 Comments
Please
log in
or
register
to add a comment.
Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.
7.3m
questions
9.7m
answers
Other Questions
How do you balance __H2SO4 + __B(OH)3 --> __B2(SO4)3 + __H2O
As an object’s temperature increases, the ____________________ at which it radiates energy increases.
Key facts of covalent bonding
How do you decide whether an observed property of matter is a physical or a chemical property?
Consider the unbalanced equation for the combustion of hexane: 2C6H14(g)+19O2(g)→12CO2(g)+14H2O(g) Determine how many moles of O2 are required to react completely with 7.9 moles C6 H14 .
Twitter
WhatsApp
Facebook
Reddit
LinkedIn
Email
Link Copied!
Copy
Search QAmmunity.org