The percent yield of the reaction is 18.9%.
In this question, we are given the mass of aluminum (5.0 g), the mass of manganese oxide (4.5 g), and the mass of aluminum oxide (1.8 g) produced at the end of the reaction. To determine the percent yield of the reaction, we need to compare the actual yield (1.8 g) to the theoretical yield, which can be calculated using stoichiometry.
The balanced chemical equation for the reaction is:
2Al + Fe2O3 → Al2O3 + 2Fe
From the equation, we can see that the molar ratio between aluminum oxide and aluminum is 1:2. Therefore, if 5.0 g of aluminum is completely reacted, the theoretical yield of aluminum oxide can be calculated as:
(5.0 g Al) x (1 mol Al/26.98 g Al) x (1 mol Al2O3/2 mol Al) x (102.0 g Al2O3/1 mol Al2O3) = 9.53 g Al2O3
The percent yield is then calculated using the formula:
Percent Yield = (Actual Yield/Theoretical Yield) x 100
Percent Yield = (1.8 g/9.53 g) x 100 = 18.9%