N2(g) + 3H2(g) 2NH3(g) + 100.4 kJ

A yield of NH3 of approximately 98% can be obtained at 200o C and 1,000 atmospheres of pressure.

What is the H in kJ of heat released per mole of NH3(g) formed?
How many grams of N2 must react to form 1.7 grams of ammonia, NH3?

N2(g) + 3H2(g) = 2NH3(g) + 100.4 kJ

A yield of NH3 of approximately 98% can be obtained at 200° C and 1,000 atmospheres of pressure.

1. What is the H in kJ of heat released per mole of NH3(g) formed?
2. How many grams of N2 must react to form 1.7 grams of ammonia, NH3?



#1 should be obvious. If it's not, you need to do some work.

100.4 kJ are released when 2 NH3 (2 moles of NH3) are formed. Therefore, if only 1 mole of NH3 is formed, 50.2 kJ of heat will be released.

#2. What is the molecular weight of N2? You have two atoms of nitrogen in one molecule of N2. The atomic weight of nitrogen is approximately 14 (you look up the exact value). Therefore, the molecular weight of nitrogen is 28.

What is the molecular weight of hydrogen? You have two atoms of hydrogen in one molecule of H2. The atomic weight of hydrogen is approximately 1. The molecular weight of hydrogen is approximately 2.

What is the molecular weight of NH3. You have one atom of nitrogen and three atoms of hydrogen. 14 + 3*1 = 17.

1.7 grams of ammonia is how many moles? To convert from grams to moles, you divide by the molecular weight

\frac{grams}{\frac{grams}{mole}} = moles

Once you know how many moles of NH3 you are producing, you can determine from the chemical equation that 2 moles of NH3 are produced from 1 mole of N2. So, divide the number of moles by 2. Now, you have to convert that back to a weight. You do that by multiplying by the molecular weight of N2.

{moles}\,{\frac{grams}{mole}} = grams

1.7 g NH3 = 1mole/ 17.03g

so grams cancel out and I have .0998 moles I don't know if that is right

.0998/2 = .0499

.0499 moles * molecular weight of N2 ( 28.02)

1.4 is that right?

1.7 g NH3 = 1mole/ 17.03g

so grams cancel out and i have .0998 moles i dont know if that is right

.0998/2 = .0499

.0499 moles * molecular weight of N2 ( 28.02)

1.4 is that right?

\frac{1.7\, g\, NH3\,}{ \frac{17.03\,g\,NH3}{1\,mole NH3}}\,= 0.9998\,moles\,NH3

0.9998 / 2 = 0.0499 moles N2 (so far, correct)

0.0499\,x\, 28.02 = 1.4075 grams of N2

Yes, it looks like you have it right. Your "1.7 g NH3 = 1mole/ 17.03g" confused me a bit, but I think you understand it correctly.

Yea I don't know how to make it look like you have it but I think I understood it. I asked another problem and I think I do it similar to this one but I'm not sure.

N2(g)+3H2(g)-->2NH3(g)+24000calories
A yield of NH3(ammonia) of approximately 98% can be obtained at 200 degrees Celsius and 1000 atm. Of pressure.
Now, how many grams of N2 must react to form 1.7 grams of NH3(ammonia)?
The answer is simpler than you think.

N2 x 28 x 47.6 N2(1.7)
---- = ---- ---- = ---- ------ or ------- = 1.4 g of N2 is the answer
NH3 1.7g 34 1.7g 34 2NH3

The atomic weight of Nitrogen is approximately 14 so N2 = 28.
The atomic number of Hydrogen is approximately 1 so...
2NH3 would be (2x14)+(1x2x3)=28+6=34

sorry layout screwed
N2/2NH3=x/1.7g
28/34=x/1.7g
47.6/1.7 (or) N2(1.7)/2NH3=1.4 grams of N2(answer)

The question was already solved and... you can make use of the edit button to edit your post.

Well, unless you're in the fatfree skin that is...

Ok, THREAD CLOSED