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## 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 N

_{2}produced, in a pression of 0.82 atm and a temperature of 27

^{o}C, 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 6NaN

_{3}= 6(23 + 3(16)) = 6(23 + 48) = 6(65) = 390 g.

Mass of Fe

_{2}O

_{3}= 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 NaN

_{3}, 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 NaN

_{3}.

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 27

^{o}C to Kelvin = 27 + 273 = 300 K.

To find the molar volume (V

_{m}): [tex]PV_m = nRT[/tex], then [tex]0.82V_m = 0.082 \times 300[/tex]; thus V

_{m}= 30 L.

Since 390 g of NaN

_{3}equal 6 mol, then the volume of N

_{2}produced is 9V

_{m}.

[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.