Answer:
14.04 L.
Explanation:
- The balanced reaction to form Al₂O₃ is:
<em>4Al + 3O₂ → 2Al₂O₃,</em>
4.0 moles of Al react with 3.0 moles O₂ to produce 2.0 moles Al₂O₃.
- Firstly, we need to calculate the no. of moles in (46.54 grams) of Al₂O₃:
<em>n = mass/molar mass </em>= (46.54 g) / (101.96 g/mol) = <em>0.4565 mol.</em>
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<u><em>Using cross multiplication:</em></u>
3.0 moles of O₂ produce → 2.0 mole Al₂O₃, from the stichiometry.
??? moles of O₂ produce → 0.4565 mole Al₂O₃.
<em>∴ The no. of moles of O₂ needed to produce 0.4565 mol (46.54 grams) of Al₂O₃</em> = (3.0 mol)(0.4565 mol)/(2.0 mol) = <em>0.6847 mol.</em>
- To calculate the volume of O₂ needed to produce 0.4565 mol (46.54 grams) of Al₂O₃, we can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 0.6847 mol).
R is the general gas constant (R = 0.082 L/atm/mol.K),
T is the temperature of the gas in K (T = 300.0 K).
<em>∴ V = nRT/P</em> = (0.6847 mol)(0.082 L/atm/mol.K)(300.0 K)/(1.2 atm) =<em> 14.04 L.</em>