Pls can someone help ASAP

1

1 mole of any gas will occupy the same volume of space (approximately 24dm® at RTP).
Given this information and the balanced equation below calculate® the volume of ammonia produced when
500 cm' of nitrogen reacts with excess hydrogen.
N(g) + 3H.(8) → 2NH,(g)
Show your working in your answer, and state® the unit.
Max. marks: 2

This is a homework question due on Monday!

Reply 1

Nitrotoluene

12

Original post

by EmileLak

1 mole of any gas will occupy the same volume of space (approximately 24dm® at RTP).
Given this information and the balanced equation below calculate® the volume of ammonia produced when
500 cm' of nitrogen reacts with excess hydrogen.
N(g) + 3H.(8) → 2NH,(g)
Show your working in your answer, and state® the unit.
Max. marks: 2
This is a homework question due on Monday!

Please note: The reaction written thus N(g) + 3H.(8) → 2NH,(g), as above, is a big mess and means nothing.
To find the volume of ammonia (NH3) produced, we need to follow the steps provided.
a) : Relate the volumes of gases using the stoichiometry below of the balanced reaction.
According to the balanced equation, 1 mole of N2 reacts to produce 2 moles of NH3. Since equal moles of gases occupy equal volumes at the same temperature and pressure, the volume ratio of the gases in the reaction is the same as their mole ratio.

Balanced reaction: N2 (g) + 3H2 (g) <==> 2NH3 (g)

Therefore, the volume of NH3 produced is twice the volume of N2 reacted.
b) : Apply the volume ratio to the given volume of nitrogen.
You have 500 cm^3 of nitrogen. Based on the stoichiometry above, this will produce twice that volume of ammonia.
-Volume of NH3 = 2 × Volume of N2
-Volume of NH3 = 2 × 500 cm^3
-Volume of NH3 = 1000 cm^3

The volume of ammonia produced is 1000 cubic centimeters (cm^3).

Note: The unit is cubic centimeters (cm^3).

My 2 cents!

(edited 11 months ago)

Reply 2

Pigster

20

Original post

by EmileLak

1 mole of any gas will occupy the same volume of space (approximately 24dm® at RTP).
Given this information and the balanced equation below calculate® the volume of ammonia produced when
500 cm' of nitrogen reacts with excess hydrogen.
N(g) + 3H.(8) → 2NH,(g)
Show your working in your answer, and state® the unit.
Max. marks: 2
This is a homework question due on Monday!

N's method is fine and I will agree with his final answer, but I will do it in a more formal way (which is longer and when you get the hang of chemical calculations you don't do - instead you will do N's method)

amount (in mol) of N2 = vol (in dm3) / 24 (dm3 as you state)
So we need to convert your stated 500 cm3 into dm3, which you do by dividing by 1000 -> 0.5 dm3
Plugging that into the above equation we get 0.5 / 24 = 0.0208 (mol of N2)

The balanced equation: N2 + 3H2 -> 2N3 shows that... if you have 1 mol of N2 it will react with 3 mol of H2 and form 2 mol of NH3. We have an excess of H2, so we can forget about that. It also means that if you have 2 mol of N2, you will be able to form 4 mol of NH3 and 10 mol of N2 can form 20 mol of NH3. If you have x mol of N2 you can form 2x mol of NH3. We happen to have 0.0208 mol of N2, so we should be able to form 2x 0.0208 mol = 0.0417 mol of NH3 (you might have expected 0.0416, but I didn't round the 0.0208 on my calculator before I hit x2)

Now all we have to do is rearrange the formula (amount = vol / 24) and get amount x 24 = vol
Plugging my amount of NH3 in, we get: 0.0417 (again the calculator value rather than this rounded version) x 24 = vol = 1 (dm3) = 1000 cm3

Pls can someone help ASAP

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