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Topic: Mole Balance for Combustion of Ammonia  (Read 4586 times)

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huma

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Mole Balance for Combustion of Ammonia
« on: January 31, 2006, 05:51:39 PM »
ok i got the first part can u help me iwth the second part?

4NH3 + 7O2 --> 4NO2 + 6H2O
a. How many grams of oxygen are necessary to produce 4.50 moles of NO2?
so i got.... 252 grams of O2
i did :
7 moles O2 produce          4 moles of NO2
x moles O2 produce          4.5 moles of NO2

x= 7 * 4.5 / 4 = 7.875 moles

m= n * MM = 7.875 * 32 = 252 grams of O2

now they ask.
b. How many molecules of water are produced when 2.25 moles of ammonia are completely reacted?

 ???

help
« Last Edit: January 31, 2006, 06:45:50 PM by geodome »

Offline Borek

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Re:help...
« Reply #1 on: January 31, 2006, 05:53:33 PM »
How many molecules of water are produced when 2.25 moles of ammonia are completely reacted?

Calculate moles - you should know how many particles are in one mol. That's just the mole definition btw.
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huma

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Re:Mole Balance for Combustion of Ammonia
« Reply #2 on: February 01, 2006, 12:27:29 PM »
ok let me get back to u on that..
first let me do the problem

shulien

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Re:Mole Balance for Combustion of Ammonia
« Reply #3 on: February 01, 2006, 11:11:11 PM »
Calculate moles - you should know how many particles are in one mol. That's just the mole definition btw.

ok let me get back to u on that..
first let me do the problem

BOREK is right.  get the mole first of water produced when 2.25 mol of ammonia completely reacted and then multiply that to value of Avogadro's number.
« Last Edit: February 01, 2006, 11:11:57 PM by shulien »

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