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Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of
dinitrogen pentoxide undergoes simple decomposition, what volume of Oz(9) is
produced at 48.0 °C and 125 kPa?


Sagot :

The volume of O₂ produced: 84.6 L

Further explanation

Given

7.93 mol of  dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

[tex]\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L[/tex]

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