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Answer :
8.6 grams of Al2O3 are formed from 15.0 liters of O2 at 97.3 kPa and 21 degrees C.
Volume of O2 = 15.0 L Pressure of O2 = 97.3 kPa ,Temperature = 21 degrees C,
Reaction: Al + O2 → Al2O3
From the reaction, we can see that one mole of Al will react with one mole of O2 to produce one mole of Al2O3.
The balanced chemical equation for the reaction is: 2Al(s) + 3O2(g) → 2Al2O3(s)
Let's use the Ideal Gas Law to calculate the moles of O2: n = PV/RT ,n(O2) = (97.3 kPa * 15.0 L) / (0.08206 L atm/mol·
K * (21°C + 273)) = 0.714 mol.
Now, let's convert the moles of O2 to moles of Al2O3 using the stoichiometric coefficients in the balanced equation.
Moles of Al2O3 = 0.714 mol O2 × 2 mol Al2O3/3 mol O2 = 0.476 mol Al2O3.
Finally, we can convert the moles of Al2O3 to grams using the molar mass of Al2O3.Mass of Al2O3 = 0.476 mol Al2O3 × 101.96 g/mol = 48.6 g Al2O3.
Therefore, 48.6 grams of Al2O3 are formed from 15.0 liters of O2 at 97.3 kPa and 21 degrees C.
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