Topic: Mole Balance for Combustion of Ammonia (Read 5030 times)

huma · « **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

Borek · « **Reply #1 on:** January 31, 2006, 05:53:33 PM »

Quote from: huma on January 31, 2006, 05:51:39 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.

huma · « **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 · « **Reply #3 on:** February 01, 2006, 11:11:11 PM »

Quote from: Borek on January 31, 2006, 05:53:33 PM

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

Quote from: huma on February 01, 2006, 12:27:29 PM

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.

Topic: Mole Balance for Combustion of Ammonia (Read 5030 times)

huma

Mole Balance for Combustion of Ammonia

Borek

Re:help...

huma

Re:Mole Balance for Combustion of Ammonia

shulien

Re:Mole Balance for Combustion of Ammonia

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