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Topic: NEED HELP ON STOICHIOMETRY!!  (Read 3390 times)

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Offline Darknight53

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NEED HELP ON STOICHIOMETRY!!
« on: January 16, 2007, 09:41:08 PM »
currently studying Stoichiometry not sure what it is but from what i know involves molar ratios, molar masses balancing and interpreting equations and conversions between grams and moles.

i really know nothing about those thing s and could really use the help
here is a problem (the first one on my practice quiz that i don't get)

N2 + 3 H2 ----> 2NH3

if anyone could explain how to get the answer i would greatly appreciate it.
« Last Edit: January 16, 2007, 09:48:08 PM by Darknight53 »

Offline enahs

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Re: NEED HELP ON STOICHIOMETRY!!
« Reply #1 on: January 16, 2007, 10:36:10 PM »
I do not understand the question? That is a balanced equation. Same number of Nitrogen on the left as on the right, and same number of Hydrogens on the left as the right.


Offline X-18

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Re: NEED HELP ON STOICHIOMETRY!!
« Reply #2 on: January 25, 2007, 12:56:31 AM »
Hello,

How the question is carried out is the same in most situations. The only main factors are what your knowns and unknowns are. The example you provided was:

N2 + 3 H2 ----> 2NH3
m=0.750g
M=28.02g/mole

but because you gave no masses or anything of the like I'm going to use my own numbers, added below the equation how I would normally do it.

For this I'm going to assume that you know the mass of N2 used in the reaction and are trying to find out the mass of NH3 produced in the reaction. The general steps used are:

1) find the number of moles of N2

n=m/M
 =0.750g/28.02g/mole
 =2.6766...x10^-2 mole N2

2) do a mole ratio

# of mole N2 * (# of mole of what you what to know / # of mole of what you know)

for this question it will be, for the second part, 2NH3 / N2 because we know how many moles there are of N2 but not NH3, also we use the mole ratio for the balanced equation

=2.6766...x10^-2 mole * 2 mole NH3   /   1 mole N2
=5.3533...x10^-2 mole NH3

3) now that we know how many moles of NH3 are produced, we can calculate the mass of NH3 produced

n=m/M
m=Mn
 =5.3533...x10^-2 mole   *   17.04 g/mole
 =0.912g

Then your done. The same general flow of it can be used for any calculation like that, regardless of your knowns and unknowns.

I'm going to cross my fingers that that is what you were wondering about, either way, that's how I took your question. Try to be more exact on what your question is.

Have fun and good luck.
X-18

Offline english

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Re: NEED HELP ON STOICHIOMETRY!!
« Reply #3 on: January 25, 2007, 01:15:31 PM »
Stoichiometry is about measuring molecular quantites by weighing them.  We cannot just count trillions of particles.  So we measure the mass of substances to calculate how many particles that quantity holds, following the derived rule that 1 mol, or 6.0223 x 1023, of particles of any element is equal to each element's atomic mass in grams.

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