If you only have 13.7 g of aluminum metal and an excess of oxygen, what is the theoretical yield of aluminum oxide (Al2O3) in grams?
A
51.8 g
B
13.7 g
C
10.2 g
D
25.9 g
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1
Start by writing the balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide: 4Al + 3O₂ → 2Al₂O₃.
Calculate the molar mass of aluminum (Al) using the periodic table. The atomic mass of Al is approximately 26.98 g/mol.
Determine the number of moles of aluminum you have by using the formula: moles of Al = mass of Al / molar mass of Al. Substitute the given mass of aluminum (13.7 g) and the molar mass you calculated.
Use the stoichiometry of the balanced equation to find the moles of aluminum oxide (Al₂O₃) that can be produced. According to the equation, 4 moles of Al produce 2 moles of Al₂O₃. Set up a ratio to find the moles of Al₂O₃.
Calculate the theoretical yield in grams of Al₂O₃ by multiplying the moles of Al₂O₃ by its molar mass. The molar mass of Al₂O₃ can be calculated as: (2 × atomic mass of Al) + (3 × atomic mass of O).