Calculate the mass of aluminum oxide (Al2O3) produced if 25.00 g of Aluminum (Al) is reacted completely.
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00 g. Given: - Molar mass of Aluminum (Al) = 26.982 g/mol - Mass of Aluminum (Al) = 25.00 g Number of moles of Aluminum (Al) = Mass / Molar mass Number of moles of Aluminum (Al) = 25.00 g / 26.982 g/mol Number of moles of Aluminum (Al) ≈ 0.926 moles Show more…
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Calculate the mass of aluminum oxide, Al2O3, formed when 12.00 g Al is reacted with 10.0 g oxygen from the air. The balanced chemical reaction is: 4Al + 3O2 → 2Al2O3 Determine the limiting reactant. Determine the theoretical yield. Determine the percent yield if you only made 2 grams of Al2O3.
Susan H.
Ma Ednelyn L.
Aluminum reacts with oxygen to give aluminum oxide. $$4 \mathrm{Al}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3(\mathrm{~s})$$ If you have $6.0 \mathrm{~mol} \mathrm{Al},$ calculate the amount (mol) and mass (g) of $\mathrm{O}_{2}$ needed for complete reaction. Calculate the mass of $\mathrm{Al}_{2} \mathrm{O}_{3},$ in grams, that is produced.
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