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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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oblivionks oblivionks
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11 years ago
I've tried this like a billion times/: and I can't get itt.

If 58 moles of NH3 are combined with 32 moles of sulfuric acid, what is the limiting reactant and how many grams of the excess reactant is left over?

Please include the work.

Thank you sooooo much (:
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#1
11 years ago
The balanced equation:
H2SO4 + 2 NH3 ----> ( NH4)2SO4
For every 2 moles of NH3 you only need 1 mole of H2SO4.
58 moles NH3 * (1 mole H2SO4 / 2 moles NH3) =
29 moles of H2SO4 used

so now 32 moles of sulfuric acid - 29 moles of sulfuric acid =
3 moles of sulfuric acid left

Now multiply by the molar mass
3 moles of H2SO4 * (98.07 grams H2SO4 / 1 mole H2SO4) =     294.21 grams of H2SO4 left over
wrote...
#2
11 years ago
The balanced reaction is:

2 NH3  +  1  H2SO4   ---->  1  (NH4)2SO4

58 mol of NH3 would produce 29 moles of (NH4)2SO4

32 mol of H2SO4 would produce 32 moles of (NH4)SO4

So, the limiting reactant would be NH3 because it produces less (NH4)2SO4<---------ANSWER

58 mol of NH3 reacts with 29 moles of H2SO4, so there would be 3 moles of H2SO4 in excess.

Grams of excess reactant = 3 moles (98 g/mol) = 294 g<---ANSWER
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