Preparing for the Chem Finals...

[Xylitol]

I have to find out the mass of water produced from 0.839 moles of ammonia (NH3)

Equation:
NH3 + 6NO -> 5N2 + 6H20

I found that there are 14 [g] of ammonia with the .839 moles

Do I find the # of moles of 6NO by multiplying .830 * 3/2?
I'm confused - I forgot a lot

 

[Brainonska511]

If you aren't worrying about limiting reagents, then You would take the number of moles of H in NH3 and then halve that number to get your moles of water. Then you do a basic moles to mass conversion.

(Moles of NH3 in experiment)*(moles of H in 1 mol of NH3)*(1 mol of H2O/2 mol H)*(mass of H20/1mol H2O)

 

[MichaelD]

See, this is why I failed chemistry. Moles are small mammals, related to rats, that live in the ground. You blow them up with explosives dropped into their holes like Chevy Chase did in Caddyshack. I know that he had a chipmunk problem...same thing essentially.

All math is identical to me. That equation you posted just hurts my head. It makes no sense at all.

NH3 + 6NO -> 5N2 + 6H20

WTF is that supposed to mean?

John has two apples. John is on a plane traveling east at 342KM/hr against a western wind blowing at 34MPH . His apples are inside a satellite in geosynchronous orbit 42.2Km above the earth. If Mary kicks John in the nuts at 23Km/hr, how many golf balls can Steve hit through an ammonium nitrate cloud at 23:02 hours on a Tuesday during a full moon. <--all math is like this

 

[Xylitol]

Originally posted by: Brainonska511
If you aren't worrying about limiting reagents, then You would take the number of moles of H in NH3 and then halve that number to get your moles of water. Then you do a basic moles to mass conversion.

(Moles of NH3 in experiment)*(moles of H in 1 mol of NH3)*(1 mol of H2O/2 mol H)*(mass of H20/1mol H2O)

what if you were worrying about limiting reagents

 

[Brainonska511]

NH3 + 6NO -> 5N2 + 6H20

WTF is that supposed to mean?

It's not that hard. It is a unbalanced equation for the formation of nitrogen gas and water from ammonia and nitrogen oxide.

It should be balanced before the stoichiometry takes place.

 

[Brainonska511]

Originally posted by: Xylitol

Originally posted by: Brainonska511
If you aren't worrying about limiting reagents, then You would take the number of moles of H in NH3 and then halve that number to get your moles of water. Then you do a basic moles to mass conversion.

(Moles of NH3 in experiment)*(moles of H in 1 mol of NH3)*(1 mol of H2O/2 mol H)*(mass of H20/1mol H2O)

what if you were worrying about limiting reagents

If you only know how much ammonia you have, then you can assume that you have excess nitrogen oxide. If it gives you a starting amount of nitrogen oxide as well, you can do a mole to mole conversion (find out which chemical will give the least amount of one product, and that is the limiting reagent).

 

[Brainonska511]

BTW, I don't know if you just mistyped it, but your equation should be:

2NH3 + 6NO --> 5N2 + 6H2O

 

[Xylitol]

Originally posted by: Brainonska511
BTW, I don't know if you just mistyped it, but your equation should be:

2NH3 + 6NO --> 5N2 + 6H2O

the question says 4NH3 - maybe its a atypo?

 

[Brainonska511]

Originally posted by: Xylitol

Originally posted by: Brainonska511
BTW, I don't know if you just mistyped it, but your equation should be:

2NH3 + 6NO --> 5N2 + 6H2O

the question says 4NH3 - maybe its a atypo?

Oops, I was the one that made the mistake. It should be 4NH3.