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

  • Thread starter pc2-brazil
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  • #1
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Thank you in advance.

Homework Statement



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).

Homework Equations



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

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.
 

Answers and Replies

  • #2
Borek
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a OK, for b I got a little bit more. Not much. Don't round down intermediate results.
 
  • #3
205
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a OK, for b I got a little bit more. Not much. Don't round down intermediate results.
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.
 
  • #4
Borek
Mentor
28,473
2,871
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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