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
1 answer:
Answer:
8.70 liters
Explanation:
First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:
- 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃
Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:
- 0.354 mol AI₂O₃ * = 0.531 mol O₂
We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:
- 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
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