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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Tonbon1221 Tonbon1221
wrote...
11 years ago
HELP!!!!! I have been stuck for a DAY!

I wrote the balanced equation:
Fe2O3 + 3Zn -----> 2Fe + 3ZnO

I found the Zn to be the limiting reactant.


how do i find how much of each product is produced?Neutral FaceNeutral Face?
how do I find how many grams of each reactant is left over?Neutral FaceNeutral Face
nvm i've found out.

just need to know exactly how to calculate the reactant left...
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wrote...
#1
11 years ago
Fe2O3 + 3 Zn ? 2 Fe + 3 ZnO

(25 g Fe2O3) / (159.6887 g Fe2O3/mol) = 0.15655 mol Fe2O3
(25 g Zn) / (65.4094 g Zn/mol) = 0.38221 mol Zn

0.15655 mole of Fe2O3 would react completely with 0.15655 x 3 = 0.46965 mole of Zn, but there is not that much Zn present, so Zn is the limiting reactant.

(0.38221 mol Zn) x (2/3) x (55.8452 g Fe/mol) = 14 g Fe produced
(0.38221 mol Zn) x (3/3) x (81.4088 g ZnO/mol) = 31 g ZnO produced
(0.38221 mol Zn) x (1/3) = 0.1274 mol Fe2O3 reacted
(0.15655 mol Fe2O3 initially - 0.1274 mol Fe2O3 reacted) x (159.6887 g Fe2O3/mol) = 4.7 g Fe2O3 left over.
There is no Zn left over.
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