Final answer:
To balance the equation between aluminum and oxygen to form aluminum oxide, we need to ensure that the number of atoms on both sides of the equation is equal. The unbalanced equation is Al + O2 → Al2O3. To balance this equation, we need to have 4 Al on the left side, which means multiplying the Al by 4. The percent yield of the reaction is 37.08%.
Step-by-step explanation:
To balance the equation between aluminum and oxygen to form aluminum oxide, we need to ensure that the number of atoms on both sides of the equation is equal. The unbalanced equation is:
Al + O2 → Al2O3
To balance this equation, we need to have 4 Al on the left side, which means multiplying the Al by 4. This gives us:
4 Al + 3 O2 → 2 Al2O3
To calculate the percent yield of the reaction, we use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100%
The actual yield given in the question is 3.50 grams of aluminum oxide. To calculate the theoretical yield, we need to convert the given mass of aluminum and oxygen to moles, and then use the balanced equation to determine the stoichiometric ratio of the reactants and products. The molar mass of aluminum is 26.98 g/mol and the molar mass of oxygen is 32.00 g/mol. The molar mass of aluminum oxide (Al2O3) is 101.96 g/mol.
Theoretical yield of Al2O3 = (2.50 g Al x 1 mol Al / 26.98 g Al) x (2 mol Al2O3 / 4 mol Al) x (101.96 g Al2O3 / 1 mol Al2O3) = 9.44 g Al2O3
Now we can calculate the percent yield:
Percent Yield = (3.50 g / 9.44 g) x 100% = 37.08%