Question

A chemist producing ammonia by the Haber process is given 10.0 L of hydrogen gas at STP to the process

a. How many moles of H2 does the chemist have?

b. How many moles of N2 will the chemist require to react with H2? (use answer a.)

c. How many grams of N2 will be required?

d. How many liters of N2 will be required at STP (use answer b.)

e. How many moles of NH3 will be produced? (use answer a.)

f. How many grams of of NH3 will be produced?

g. How many liters of NH3 will be produced at STP? (use answer e.)

Answer 1

@Photon336

Answer 2

@Zale101

Answer 3

for a. I got 0.446 mole of H2 gas

Answer 4

i am stuck on b.

Answer 5

i don't think I need to use pv=nrt

Answer 6

\(\Large N_{2(g)}+3H_{2(g)} \rightarrow 2NH_{3(g)}\)
After you get the moles of H2, use mole to mole ration to find the moles for N2

Answer 7

mole to mole ratio*

Answer 8

Are you familiar with mole to mole ratio?

Answer 9

No not super familar can you show me plz

Answer 10

this isn't homework just practice

Answer 11

i thought when reacting H2 and N2 it is already balanced

Answer 12

Sure.
You said you got 0.446 mole of H2 gas. We know that for every 1 mole of N2, there's 3 moles of H2.

So, \( 0.446 ~mole~ H_2~\times \Large \frac{1~mole~N_2}{3~mole~H_2} \)
Moles of H2 will cancel out and that will leave you with moles of N2

Answer 13

oh nvm about the balancing the equation I see that ammonia is stated in the problem

Answer 14

Okay let me look at this and study it

Answer 15

okay I am going to try to anser the rest myself if I have any further questions I will tag you thanks

Answer 16

No problem.

Answer 17

lol so how do we solve part c?

Answer 18

wait... actually let me do this

Answer 19

okay how do we solve part e

Answer 20

Okay.
Question C says: c. How many grams of N2 will be required?
Do you remember how to convert moles to grams?

Answer 21

e. How many moles of NH3 will be produced? (use answer a.)

Same thing with B, you do mole to mole ratio.

Answer 22

i got c

Answer 23

There's 3 moles of H2 for every 2 moles of NH3

Answer 24

and d. how do we get part e using a.

Answer 25

okay i agree with your statement but why can't we use N2 instead of 3H2

Answer 26

are we going to be using a mole to mole ratio for part e?

Answer 27

do you know what I mean?

Answer 28

we are using H2 because we want to see how many moles of NH3 will be produced after the reaction?

Answer 29

so this is the reason we use H2 instead?

Answer 30

zale?

Answer 31

Actually, you can use whatever moles whether it's N2 or H2, but your question suggests to use answer a.

Answer 32

Sorry for the late reply. I was studying too.

Answer 33

oh okay

Answer 34

no problem I thought maybe you were having computer malfunctions lol

Answer 35

lol

Answer 36

h. suppose the chemist finds that they have on hand 3.00 L of nitrogen gas. Is this more or less than the amount required to react with the 10.0 L of hydrogen?

i. Calculate the volume (in L) of ammonia at STP that could be produced from the 3.00 L of nitrogen in part h.

j. Looking at your calculations above, determine which produces more ammonia: 10.0 L of hydrogen gas or 3.00 L of nitrogen gas ? The reactant that produces the least amount of product is called the limiting reactant for the reaction. Of course, this depends on the actual amount of each reactant you have to start with, not simply on the stoichiometry of the balanced equation.

Answer 37

@Zale101

Answer 38

At STP that translates to \(\large \frac{22.4~L}{1~mole}\)