Chem calculation question help needed

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

Reply 1

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.

Reply 2

coconut64

Original post by alow

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

Reply 3

alow

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?

Reply 4

coconut64

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

Reply 5

alow

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.

Reply 6

coconut64

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?

Reply 7

alow

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.

Reply 8

coconut64

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?

Reply 9

alow

Original post by coconut64

Right, so I got 200cm3 what do I do next?

You should have 200cm³ of ammonia, plus 100cm³ of nitrogen.

Reply 10

coconut64

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!

Reply 11

alow

Original post by coconut64

Okay that makes sense. I thought it was just asking for the volume of ammonia but nope. Thanks!

No problem :smile:

Reply 12

coconut64

Original post by alow

No problem :smile:

Just wondering do you help with biology by any chance? Thx

Reply 13

alow

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.

Reply 14

coconut64

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?

Reply 15

alow

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.

Reply 16

coconut64

Original post by alow

Yeah I'm good at enzymes stuff.

Great! Please help me with this https://www.reddit.com/r/HomeworkHelp/comments/51akb9/high_school_biology_effect_of_temperature_on_the/

Reply 17

alow

Original post by coconut64

Great! Please help me with this https://www.reddit.com/r/HomeworkHelp/comments/51akb9/high_school_biology_effect_of_temperature_on_the/

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.