excess aluminum reacts with 3.50 moles oxygen to form aluminum oxide. the reaction generated 105 g al₂o₃…

Explanation:

Step1: Determine moles of $Al_2O_3$ from oxygen

From the balanced - equation $4Al + 3O_2\rightarrow2Al_2O_3$, the mole ratio of $O_2$ to $Al_2O_3$ is $3:2$. Given $n_{O_2}=3.50$ moles. $n_{Al_2O_3,theoretical}=\frac{2}{3}\times n_{O_2}=\frac{2}{3}\times3.50\ mol=\frac{7}{3}\ mol$

Step2: Calculate the theoretical mass of $Al_2O_3$

The molar mass of $Al_2O_3$ is $M=(2\times27 + 3\times16)\ g/mol=(54 + 48)\ g/mol = 102\ g/mol$. $m_{Al_2O_3,theoretical}=n_{Al_2O_3,theoretical}\times M=\frac{7}{3}\ mol\times102\ g/mol = 238\ g$

Step3: Calculate the percent - yield

The percent - yield formula is $\text{Percent Yield}=\frac{m_{actual}}{m_{theoretical}}\times100%$. Given $m_{actual}=105\ g$ and $m_{theoretical}=238\ g$. $\text{Percent Yield}=\frac{105\ g}{238\ g}\times100%\approx44.1%$

Answer:

$44.1$