During a laboratory experiment, 36.12 grams of Aboz was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What was the volu

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During a laboratory experiment, 36.12 grams of Aboz was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What was the volume of O2
used during the experiment?
30 2+ 4A1 — 2AI2O3
8.70 liters
9.22 liters
10.13 liters
12.81 liters

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  1. Answer:

    8.70 liters

    Explanation:

    • 3O₂+ 4Al → 2AI₂O₃

    First we convert 36.12 g of AI₂O₃ into moles, using its molar mass:

    • 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃

    Then we convert AI₂O₃ moles into O₂ moles, using the stoichiometric coefficients of the reaction:

    • 0.354 mol AI₂O₃ * [tex]\frac{3molO_2}{2molAl_2O_3}[/tex] = 0.531 mol O₂

    We can now use the PV=nRT equation to calculate the volume, V:

    • 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
    • V = 8.708 L
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