What volume of N2 (in dm3) is produced from 140 g of NaN3....

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Discussion Overview

The discussion revolves around calculating the volume of nitrogen gas (N2) produced from the decomposition of sodium azide (NaN3) in a motor vehicle safety airbag. Participants explore the stoichiometry of the reactions involved and the appropriate calculations to determine the volume of gas produced under specific conditions (273K and 1 x 10^5 Pa).

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant questions whether the volume of N2 is simply 22.7 dm3, suggesting it might be a trick question.
  • Another participant clarifies that the molar volume refers to one mole, not the total produced, and confirms that 3 moles of N2 are produced from 2 moles of NaN3.
  • A participant calculates that there are 2.158 moles of NaN3 and asserts that this leads to 3 moles of nitrogen from the first reaction.
  • Disagreement arises regarding the interpretation of stoichiometric coefficients, with one participant correcting another about the actual moles of Na produced and the subsequent nitrogen yield.
  • Participants discuss the need to account for both reactions to find the total nitrogen produced, with calculations leading to different intermediate values.
  • One participant suggests a method of combining the reactions to simplify the calculation, proposing a net reaction approach to determine the total moles of N2 produced.

Areas of Agreement / Disagreement

There is no consensus on the final volume of N2 produced, as participants present differing calculations and interpretations of stoichiometric relationships. Some participants agree on the need to consider both reactions, while others express confusion over the correct application of stoichiometry.

Contextual Notes

Participants express uncertainty about the correct application of stoichiometric coefficients and the relationship between moles of reactants and products. There are unresolved calculations regarding the total volume of nitrogen produced, and assumptions about the reactions are not uniformly accepted.

Who May Find This Useful

This discussion may be useful for students or individuals interested in chemical reactions, stoichiometry, and gas laws, particularly in the context of practical applications like airbag deployment.

dopeantelope
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Please post this type of questions in HW section using the template.
A motor vehicle safety air bag contains sodium azide (NaN3) and potassium nitrate (KNO3). On impact, N2 is produced by the successive reactions:

Reaction 1: 2NaN3—> 2Na + 3N2
Reaction 2: 10Na + 2KNO3—> K2O + 5Na2O + N2

What volume of N2 (in dm3) is produced from 140 g of NaN3 at 273K and 1 x 105 Pa pressure?

Pay attention to significant figures as you calculate your answer and give your answer as a numericla figure only.

(Take Ar Na = 23.0, N =14.0; vol. 1 mol N2 = 22.7dm3 at 105Pa, 273 K).
____________________________________

I'm not looking for someone to give me the actual answer, I just would like someone to point me in the right direction and guide me please :)

I wasn't sure if this was a trick question at first, as there is only one mole of N2 as the final product is it just 22.7dm3?

Then I wasn't sure if it was just in the first equation, as there is 3 moles of N2, would i just multiply 22.7 by 3?

I'm not sure.

Any help/guidance would be greatly appreciated, thank you in advance! :)
 
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How many moles of the azide are initially present? How many moles of nitrogen are produced in the first step? After the first step, how many moles of Na are ready for the second step?

Molar volume is by definition volume of one mole, it doesn't mean you have one mole of nitrogen produced.

dopeantelope said:
if it was just in the first equation, as there is 3 moles of N2, would i just multiply 22.7 by 3?

Yes.
 
There is 2.158 moles of NaN3 initially.. which i calculated by dividing grams by the Ar. (could i just look at the equation for this value?) 3 moles of nitrogen produced in the first step. and 10 moles of Na ready for the second step. I am just using the equation for these values. But according to the equation there is one mole of nitrogen produced in the final step, so is the answer just 22.7? @Borek
 
okay i see, so its just the ratio then. so now I've got 3.231 moles of nitrogen produced in the first step. do i multiply this by the 22.7 and get my final answer? @Borek
 
Nope, it is not the final answer yet. There is a second reaction.
 
Okay so according to the first equation, there are also 2.1538 mol of Na.
According to equation 2 there are 10 moles of Na. So i multiplied 2.1538 by 5 to get 10.769 moles of Na in the 2nd equation.

I then divided this by 10 to get 1.0769 moles of nitrogen.

So I multiplied 1.0769 by 22.7 to get 24.4dm3? @Borek
 
dopeantelope said:
According to equation 2 there are 10 moles of Na.

No. You are still mistaking stoichiometric coefficients (which tell you what is the ratio of substances reacting and/or being produced) with the real amount of the substance.

There are no 10 moles of Na. There are 2.154 moles of Na.
 
Sorry, so there are 2.154 moles of Na. So i divide this by 10 to get the number of moles of nitrogen?

0.2154 moles of nitrogen x 22.7 = 4.89dm3? @Borek
 
  • #10
Technically you divide by 10 and multiply by 1, you have to take both stoichiometric coefficients into account.

So what is the final answer?
 
  • #11
2.154 divided by 10, multiplied by 1 to give 0.2154 moles of nitrogen.
as 1 mol of nitrogen = 22.7dm3

0.2145 x 22.7dm3 = 4.87dm3 @Borek
 
  • #12
Haven't you forgot about something? See post #5.
 
  • #13
oh i see, so i need to add both together.

3.231 x 22.7 = 73.3437
0.2145 x 22.7 = 4.86915

73.3437 + 4.86915 = 78.2 dm3 @Borek
 
  • #14
Looks OK.
 
  • #15
thank you so much for your help! @Borek
 
  • #16
Why not just multiply the 1st reaction by 5, add the two reactions together and use the net rxn as the standard.
Net: 10NaN3 +2KNO3 => K2O + 5N2O + 16N2
140g/65 = 2.154 mole NaN3 => [10 / 2.154 = 16 / x] => x = 3.44 mole N2(g)
Vol = 3.44(22.7)dm3 = 78.2dm3
Just a suggestion
 

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