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# Chem calculation question help needed watch

1. In a pilot plant making ammonia, NH3, 200 cm3 of nitrogen are mixed with 300 cm3 of
hydrogen.
What would be the final volume (at the same temperature and pressure) if complete reaction occurs?

A 200 cm3
B 250 cm3
C 300 cm3
D 400 cm3
Can anyone explain how this can be worked out without moles ?
Cheers
2. Write the equation for the reaction.

Assume the same volume of gas has the same number of molecules for any approximately ideal gas.

Work out which reagent is in excess and how much will be left over.

The rest of the reactants will give ammonia.
3. (Original post by alow)
Write the equation for the reaction.

Assume the same volume of gas has the same number of molecules for any approximately ideal gas.

Work out which reagent is in excess and how much will be left over.

The rest of the reactants will give ammonia.
I worked out that N2 is in excess, so I used the n of H2 /3 then *2 then *24000; this gives me 200 which is not correct
4. (Original post by coconut64)
I worked out that N2 is in excess, so I used the n of H2 /3 then *2 then *24000; this gives me 200 which is not correct
Rememer it's:

N2 + 3H2 --> 2NH3

So you use 100cm3 of nitrogen and 300cm3 of hydrogen to make how much ammonia?
5. (Original post by alow)
Rememer it's:

N2 + 3H2 --> 2NH3

So you use 100cm3 of nitrogen and 300cm3 of hydrogen to make how much ammonia?
The equation is in fact given in the question; I worked out 200cm3 are made
6. (Original post by coconut64)
The equation is in fact given in the question; I worked out 200cm3 are made
Yeah that's right. You still have some nitrogen left over though.
7. (Original post by alow)
Yeah that's right. You still have some nitrogen left over though.
But that much of nitrogen is in excess though so there wouldn't be enough hydrogen to react to it. How does that work?
8. (Original post by coconut64)
But that much of nitrogen is in excess though so there wouldn't be enough hydrogen to react to it. How does that work?
Nothing happens to it. It's still got 100cm3 of volume though.
9. (Original post by alow)
Nothing happens to it. It's still got 100cm3 of volume though.
Right, so I got 200cm3 what do I do next?
10. (Original post by coconut64)
Right, so I got 200cm3 what do I do next?
You should have 200cm3 of ammonia, plus 100cm3 of nitrogen.
11. (Original post by alow)
You should have 200cm3 of ammonia, plus 100cm3 of nitrogen.
Okay that makes sense. I thought it was just asking for the volume of ammonia but nope. Thanks!
12. (Original post by coconut64)
Okay that makes sense. I thought it was just asking for the volume of ammonia but nope. Thanks!
No problem
13. (Original post by alow)
No problem
Just wondering do you help with biology by any chance? Thx
14. (Original post by coconut64)
Just wondering do you help with biology by any chance? Thx
I can with biochem/cell biology/genetics.

Anything else I'll try, but I'm a bit rusty.
15. (Original post by alow)
I can with biochem/cell biology/genetics.

Anything else I'll try, but I'm a bit rusty.
Any help would be appreciated since you seem quite knowledgeable. Are you okay with enzymes stuff?
16. (Original post by coconut64)
Any help would be appreciated since you seem quite knowledgeable. Are you okay with enzymes stuff?
Yeah I'm good at enzymes stuff.
17. (Original post by alow)
Yeah I'm good at enzymes stuff.
18. (Original post by coconut64)
I would think it's just that the rate of reaction is limited by the E.S --> E + P step (the step going from enzyme-substrate complex to the free enzyme and product) and at a certain concentration of substrate all of the enzyme is saturated so the increased number of collisions doesn't have an effect as there are no empty active sites.

This is all obviously assuming the enzyme doesn't begin to denature at these temperatures.
19. (Original post by alow)
I would think it's just that the rate of reaction is limited by the E.S --> E + P step (the step going from enzyme-substrate complex to the free enzyme and product) and at a certain concentration of substrate all of the enzyme is saturated so the increased number of collisions doesn't have an effect as there are no empty active sites.

This is all obviously assuming the enzyme doesn't begin to denature at these temperatures.
It took me a while to understand it but thanks.

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