How Much Gas is Produced from 8.8g of N2O Decomposition at STP?

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Discussion Overview

The discussion revolves around the decomposition of dinitrogen monoxide (N2O) and the calculation of the volume of gases produced at standard temperature and pressure (STP) from 8.8 grams of N2O. Participants engage in solving a homework problem that involves stoichiometry and gas laws.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the initial calculation for the volume of nitrogen produced, suggesting a method based on the molar mass of nitrogen and the decomposition reaction.
  • Another participant challenges the arithmetic in the first calculation, indicating a potential error in the division of mass by molar mass.
  • A different approach is proposed, where the participant suggests using a different stoichiometric interpretation of the reaction, leading to a different mass and mole calculation for nitrogen.
  • There is a correction regarding the balanced chemical equation, with a participant stating that the correct reaction should involve 2 moles of N2O producing 2 moles of N2 and 1 mole of O2.
  • Some participants express confusion about whether to use the molar mass of nitrogen as 14 g/mol or 28 g/mol, leading to different calculations for the number of moles and corresponding gas volumes.
  • One participant calculates the volume of oxygen produced and combines it with the volume of nitrogen to find a total volume, but this calculation is also questioned by others.
  • Several participants point out arithmetic mistakes in previous posts, indicating a need for careful verification of calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the calculations, with multiple competing views on the stoichiometry and the correct molar masses to use. The discussion remains unresolved with ongoing corrections and challenges to earlier claims.

Contextual Notes

There are limitations in the clarity of the stoichiometric relationships and the assumptions regarding the molar masses of nitrogen and oxygen. The calculations depend on the correct interpretation of the balanced chemical equation and the proper application of gas laws.

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Homework Statement


Dinitrogen monoxide (nitrous oxide) N2O decomposes producing nitrogen and oxygen gases.
The atomic masses are N: 14.0 and O: 16.0
The molar volume of an ideal gas at STP is 22.4 dm3/mol

Homework Equations


Calculate the total volume of produced gases at STP, when 8.8 grams of dinitrogen monoxide decomposes?

The Attempt at a Solution


Vtotal des gaz=VN+VO

N2O -------> 2N + O
8.8g---------mN
44g/mol------14g/mol

mN=(8.8*14)/44=2.8g
nN=2.8/14=0.5mol.

22.4=VN/nN
VN=22.4*nN
VN=22.4*0.5=11.2L?
 
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2,8/14 =/= 0.5. Make the division again.

OK, what's V_O equal with ? Then just add the results.
 
It shouldn't be like this ?
N2O -------> 2N + O
8.8g---------mN
44g----------2*14g

then mN=5.6g.
nN=mN/MN=5.6/14=0.4mol

22.4=VN/nN
VN=22.4*0.4=8.96L

Please tell me if this is correct, I'm confused if we should put 2*14 in this method or not. and then when calculating n, i used only 14.

if it's correct, we do same for O and then we add the volumes to each other to find the total volumes of gases.
 
Last edited:
The correct reaction should be of course

2\mbox{N}_{2}\mbox{O} \longrightarrow 2\mbox{N}_{2} + \mbox{O}_{2}

Apologies, I didn't check your equation for correction and jumped directly into the numbers.

Please, redo the calculations.

Thanks
 
2N2O-------> 2N2+O2
8.8g----------mN
2*44g--------2*28

mN=(2*28*8.8)/88=5.6g

nN=mN/MN
nN=5.6/14 (im not sure if we should divide by 14 or by (2*14))
nN=0.4mol

22.4=VN/nN
VN=22.4*0.4
VN=8.96L
 
Last edited:
No, no. Because of the molecule being diatomic, the molar mass of nitrogen is 2x14 = 28 grams/mol. So it's 0.2 moles of nitrogen = 4.48 liters of nitrogen normal cond of temp and pressure.
 
Last edited:
I don't get it, i just finished editing my previous solution, please check it!
 
chawki said:
2N2O-------> 2N2+O2
8.8g----------mN
2*44g--------2*28

mN=(2*28*8.8)/88=5.6g

nN=mN/MN
nN=5.6/14 (im not sure if we should divide by 14 or by (2*14))
nN=0.4mol

22.4=VN/nN
VN=22.4*0.4
VN=8.96L

The issue is in the bolded part. The number of moles of nitrogen is 5.6 grams/28 grams/mol = 0.2 mols.

So the volume which corresponds is 0.2 x 22.4 = 4.48l
 
ok so for O2

2N2O----------->2N2+O2
8.8g-----------------mO2
2*44g---------------32g

mO2= 3.2g
nO2=mO2/MO2=3.2/32=0.1 mol of O2

22.4=VO2/nO2
VO2=2.24L

Vtotal=4.48+2.24=6.72L

should we write mN or mN2
same thing for O2
 
Last edited:
  • #10
dextercioby said:
No, no. Because of the molecule being diatomic, the molar mass of nitrogen is 2x14 = 28 grams/mol. So it's 0.2 moles of nitrogen = 4.48 liters of nitrogen normal cond of temp and pressure.

The 2*14 comes from N2 or from 2N, i guess it's from N2!
 
  • #11
You need to check your arithmetics. The numbers in post #9 are wrong.
 
  • #12
Yes it was a terrible mistake, please check the post #9
 
  • #14
Thank you!
 

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