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

  1. Oct 5, 2015 #1
    • 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!! :)
     
  2. jcsd
  3. Oct 5, 2015 #2

    Borek

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    Staff: Mentor

    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.

    Yes.
     
  4. Oct 5, 2015 #3
    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. Im 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
     
  5. Oct 5, 2015 #4

    Borek

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  6. Oct 5, 2015 #5
    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
     
  7. Oct 5, 2015 #6

    Borek

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    Nope, it is not the final answer yet. There is a second reaction.
     
  8. Oct 5, 2015 #7
    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
     
  9. Oct 5, 2015 #8

    Borek

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    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.
     
  10. Oct 5, 2015 #9
    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
     
  11. Oct 5, 2015 #10

    Borek

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    Technically you divide by 10 and multiply by 1, you have to take both stoichiometric coefficients into account.

    So what is the final answer?
     
  12. Oct 5, 2015 #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
     
  13. Oct 5, 2015 #12

    Borek

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    Haven't you forgot about something? See post #5.
     
  14. Oct 5, 2015 #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
     
  15. Oct 5, 2015 #14

    Borek

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    Looks OK.
     
  16. Oct 5, 2015 #15
    thank you so much for your help! @Borek
     
  17. Oct 5, 2015 #16

    James Pelezo

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    Gold Member

    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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