Answer to Question #12739 in Chemistry for Molly

Question #12739
What is the mass of iron(III) bromide (295.55 g/mol) that yields 0.188 g of silver bromide (187.77 g/mol) precipitate?

__FeBr3(s) + __AgNO3(aq) → __AgBr(s) + __Fe(NO3)3(aq)
1
Expert's answer
2012-09-14T09:04:40-0400
m 0.188
FeBr3(s) + 3AgNO3(aq) = 3AgBr(s) + Fe(NO3)3(aq)
295.55 3*187.77

m = 295.55*0.188/(3*187.77) = 0.986 g

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
APPROVED BY CLIENTS