Answer :
0.000459 moles of Al₂O₃ are formed from 15.0 L of O₂ at 97.3 kPa and 21°C.
The balanced chemical equation for the formation of Al₂O₃ from O₂ is:
4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
According to the stoichiometry of the equation, 3 moles of O₂ react with 4 moles of Al to form 2 moles of Al₂O₃.
To determine the number of moles of Al₂O₃ formed from 15.0 L of O₂ at 97.3 kPa and 21°C, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We first need to convert the volume from liters to cubic meters, since the units of pressure in the ideal gas law are in pascals, which are derived from SI units:
15.0 L = 0.015 m^3
Next, we need to convert the temperature from Celsius to Kelvin:
T = 21°C + 273.15 = 294.15 K
We can now solve for the number of moles of O₂ using the ideal gas law, assuming that O₂ behaves as an ideal gas at the given conditions:
n(O₂) = PV/RT = (97.3 kPa)(0.015 m³)/(8.31 J/(mol·K) × 294.15 K) = 0.000689 mol
Finally, using the stoichiometry of the balanced chemical equation, we can determine the number of moles of Al2O₃ formed:
n(Al₂O₃) = (2/3) × n(O₂) = (2/3) × 0.000689 mol = 0.000459 mol
Therefore, 0.000459 moles of Al₂O₃ are formed from 15.0 L of O₂ at 97.3 kPa and 21°C.
To learn more about stoichiometry here
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