Final answer:
The percent yield of the reaction between 56.3g of Al and excess O₂ that generates 22.7g of Al₂O₃ is 21.4%.
Step-by-step explanation:
The percent yield of a reaction is a measure of how efficiently the reactants are converted into products. It is calculated by dividing the actual yield by the theoretical yield, and then multiplying by 100%. In this case, the actual yield of Al₂O₃ is 22.7g. To find the theoretical yield, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction. In this reaction, aluminum (Al) is the limiting reactant because it is present in the smallest amount. Using the molar masses of Al (26.98 g/mol) and Al₂O₃ (101.96 g/mol), we can calculate the theoretical yield of Al₂O₃. The molar ratio between Al and Al₂O₃ is 2:1, meaning that for every 2 moles of Al, we will get 1 mole of Al₂O₃. Therefore, we can calculate the moles of Al₂O₃ that would be produced if all of the Al reacted. Finally, we can convert the moles of Al₂O₃ to grams using its molar mass, which gives us the theoretical yield of Al₂O₃. The percent yield is then calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. Therefore, the percent yield in this reaction is:
Percent yield = (actual yield / theoretical yield) x 100%
Percent yield = (22.7g / theoretical yield) x 100%
To find the theoretical yield, we first need to calculate the moles of Al₂O₃ that would be produced if all of the Al reacted:
Moles of Al = (56.3g / 26.98 g/mol) = 2.087 mol
Moles of Al₂O₃ = (2.087 mol / 2) = 1.044 mol
Finally, we can convert the moles of Al₂O₃ to grams using its molar mass:
Theoretical yield = (1.044 mol) x (101.96 g/mol) = 106g
Now we can calculate the percent yield:
Percent yield = (22.7g / 106g) x 100% = 21.4%