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Stoichiometry: finding the yield in reaction of sodium azide with iron oxide

  1. Nov 14, 2008 #1
    Thank you in advance.

    1. The problem statement, all variables and given/known data

    Nitrogen gas can be obtained from the reaction between sodium azide and iron oxide.

    [tex]6NaN_3_{(s)}+Fe_2O_3_{(s)} \rightarrow 3Na_2O_{(s)}+2Fe_{(s)}+9N_2_{(g)}[/tex]

    a) What is the yield of this reaction, knowing that, from 390 g of azide and 400 g of iron oxide, 100 g of metalic iron were produced?

    b) What is the volume (in litres = L) of N2 produced, in a pression of 0.82 atm and a temperature of 27oC, in the conditions given in item a?

    Universal constant of gases = R = 0.082 (atm L)/(mol K).

    2. Relevant equations

    [tex]PV = nRT[/tex] (P = atm, V = litres, n = mol, R = universal constant of gases, T = Kelvin).

    3. The attempt at a solution

    a) First, the molar relation between sodium azide and iron oxide, in order to discover what is the limiting reactant:

    Mass of 6NaN3 = 6(23 + 3(16)) = 6(23 + 48) = 6(65) = 390 g.
    Mass of Fe2O3 = 2(56) + 3(16) = 112 + 48 = 160 g.
    Mass of 2Fe = 2(56) = 112 g.
    [tex]6 mol NaN_3 \rightarrow 1 mol Fe_2O_3[/tex]
    [tex]390 g NaN_3\rightarrow 160 g Fe_2O_3[/tex]

    Since we already have 390 g of NaN3, a rule of three will not be necessary.
    This mass of sodium azide produces 160 g of iron oxide; thus, there is excess of iron oxide. Therefore, the limiting reactant is NaN3.

    Mass of metalic iron produced:
    [tex]6 mol NaN_3 \rightarrow 2 mol Fe[/tex]
    [tex]390 g \rightarrow 112 g[/tex]
    The mass of metalic iron is 112g. Since the problem states the mass produced is 100 g, dividing 100 / 112 will give the yield (R):
    [tex]R = \frac{100}{112}[/tex]
    R equals approximately 89.2%.

    b) Convert 27oC to Kelvin = 27 + 273 = 300 K.
    To find the molar volume (Vm): [tex]PV_m = nRT[/tex], then [tex]0.82V_m = 0.082 \times 300[/tex]; thus Vm = 30 L.
    Since 390 g of NaN3 equal 6 mol, then the volume of N2 produced is 9Vm.
    [tex]V = 9V_m = 9 \times 30 = 270 L[/tex]
    But the yield is 89.2%:
    [tex]V = 0.892 \times 270[/tex]
    V equals 240.82 L.
     
  2. jcsd
  3. Nov 14, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    a OK, for b I got a little bit more. Not much. Don't round down intermediate results.
     
  4. Nov 15, 2008 #3
    Thank you for the response. This calculation was done by hand (100/112 = 50/58 = 25/28 = ~89,2). The next digit would be 8, which makes difference in the result (270 * 89.28 = 271.056). Anyway, if we were to do the calculation of the volume like this:
    [tex]V = \frac{100 \times 270}{112} = \frac{25 * 270}{28} = \frac{25 * 135}{14}[/tex]
    the result would be better.
    It is very good to know that the reasoning to solve the problem is right.
     
  5. Nov 15, 2008 #4

    Borek

    User Avatar

    Staff: Mentor

    It is possible that part of the difference is in molar masses, I am using my stoichiometry calculator and it does all calculations using as exact molar masses as possible.
     
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