What mass of nitrogen Dioxide, NO2, would be produced by the decomposition of 75.0g of copper (II) nitrate?
2 answers:
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Explanation:
The given data is as follows.
Pressure (P) = 760 torr = 1 atm
Volume (V) = = 0.720 L
Temperature (T) = = (25 + 273) K = 298 K
Using ideal gas equation, we will calculate the number of moles as follows.
PV = nRT
Total atoms present (n) =
=
= 0.0294 mol
Let us assume that there are x mol of Ar and y mol of Xe.
Hence, total number of moles will be as follows.
x + y = 0.0294
Also, 40x + 131y = 2.966
x = 0.0097 mol
y = (0.0294 - 0.0097)
= 0.0197 mol
Therefore, mole fraction will be calculated as follows.
Mol fraction of Xe =
=
= 0.67
Therefore, the mole fraction of Xe is 0.67.