3) consider the following reaction, 4al + 3o₂ → 2al₂o₃. julia combines 15 grams of aluminum with excess…

Explanation:

Step1: Calculate moles of Al

The molar mass of Al is 26.98 g/mol. Moles of Al = $\frac{mass}{molar - mass}$. So, moles of Al = $\frac{15\ g}{26.98\ g/mol}\approx0.556\ mol$.

Step2: Determine moles of $Al_2O_3$ from stoichiometry

From the balanced equation $4Al + 3O_2\rightarrow2Al_2O_3$, the mole - ratio of Al to $Al_2O_3$ is 4:2 or 2:1. So, moles of $Al_2O_3$ produced if all Al reacts (theoretical yield) = $\frac{0.556\ mol}{2}= 0.278\ mol$.

Step3: Calculate the theoretical mass of $Al_2O_3$

The molar mass of $Al_2O_3$ is $2\times26.98\ g/mol+3\times16.00\ g/mol = 101.96\ g/mol$. Theoretical mass of $Al_2O_3$ = moles $\times$ molar mass = $0.278\ mol\times101.96\ g/mol\approx28.35\ g$.

Step4: Calculate percent yield

Percent yield = $\frac{actual\ yield}{theoretical\ yield}\times100%$. Given actual yield = 25 g and theoretical yield = 28.35 g. Percent yield = $\frac{25\ g}{28.35\ g}\times100%\approx88.2%$.

Answer:

88.2%