Verified answer

NH3 is one nitrogen atom and 3 hydrogen atoms. N has a mass of 14 units, hydrogen has a mass of 1 unit (and there are 3 of them), so NH3 has a mass of 17. This means NH3 has 14 parts nitrogen to 3 parts hydrogen, by mass, 14:3. If the 3 H's is 3.5 grams, then there would have to be 42.5 g of Nitrogen to keep this ratio. Added up, there would be 45.5g of NH3. Hope I didn't take you in circles.

The balanced equation is:

N2 + 3H2 = 2NH3

The moles of H2 gas present is 3.5 grams / 2.0 grams/mole = 1.75 moles

Note from the equation that 3 moles of H2 produces 2 moles of ammonia.

Therefore the moles of NH3 produce is 3/2 (or 1.5) times the number of moles of hydrogen gas.

Moles NH3 = 1.5 X 1.75 moles = 2.625 moles

And since the molecular mass of NH3 is 17 grams/mole, the number of grams of ammonia produced will be:

(17 grams/ mole) X 2.625 moles = 44.625 grams

But it is best to express the answer as 45 grams, since only two significant figures were given in the data for the problem.

(Ignore all of the periods in the equations, they are just so that the spacing hopefully works out right)

Ok, stoichiometry is all about ratios, so you will set up a series of ratios that take you from 3.5g hydrogen gas to Xg NH3. To start, you need to figure out how many moles of hydrogen gas you have. To do this, figure out the molar weight of hydrogen gas (remember that hydrogen gas is diatomic, so it is written as H2). The molar weight of the element hydrogen is 1.0079 g/mol, so H2 would be double it, or 2.0158 g/mol. The first part of the ratio would look like this:

3.5g H2.....1 mol H2

----------- x -----------------

...............2.0158g H2

The second step is to figure out how many moles of the element hydrogen are present in that amount of H2 gas. You dont have to do any math yet, but you have to add a simple ratio onto the end of it that does that, the whole thing will now look like this:

3.5g H2.....1 mol H2.....2 mol H

----------- x ---------------- x -------------

...............2.0158g H2...1 mol H2

Now that you know how many moles of hydrogen element you have, you need to figure out how many moles of hydrogen gas are in one mole of NH3. If you look at the formula for NH3, it is clear that 1 molecule of NH3 contains 3 atoms of H, so one mole of NH3 would contain 3 moles of H. Shown in a ratio, it now looks like this:

3.5g H2.....1 mol H2......2 mol H....1 mol NH3

----------- x ---------------- x ------------ x --------------

...............2.0158g H2...1 mol H2...3 mol H

Now that we have the moles of NH3, we just need to figure out the molar weight of NH3. Since H has a weight of 1.0079 g/mol, and there are three there, H accounts for 3.0237 grams in NH3. The molar weight of N is 14.006 g/mol, so when you add the two, you get 17.0297 g/mol. That means that one mole of NH3 weighs 17.0297 grams. To put this into a ratio form, you add this to the equation:

1 mol NH3....17.0297g NH3

--------------- x --------------------

3 mol H.........1 mol NH3

I didnt add that to the whole equation because it is too long, but if you do, you can see that every unit (ex. g H2, mol NH3)

on top is cancelled out by one below, leaving only the g NH3 at the end. All you have to do now is put 3.5 into your calculator and go down the line, if a number is on the bottom, you divide by it, if it is on the top, you multiply. If I did it right, the answer should be 19.7122 g NH3.

If you have any trouble, feel free to email me at [email protected] and I'll do my best to help ya!

3H2 + N2 ---> 2NH3 or more simply, for each 3 moles of H you get one mole of NH3. Weight of H is 1.008 (3 moles weigh 3.024); weight of NH3 is 14.007 + 1.008 + 1.008 + 1.008 = 17.031 (I'm not using H2 so as not to make it more complicated.)

Set up an algebraic ratio. If 3.024 g H yields 17.031 g NH3, then how much (X) does 3.5 g yield?

X/17.031 g NH3 = 3.5 g H/3.024 g H

3.024 X = 3.5( 17.031 g NH3)

X = 3.5(17.031 g NH3)/3.024

X =19.712 g NH3