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

excess aluminum reacts with 3.50 moles oxygen to form aluminum oxide. the reaction generated 105 g al₂o₃. what is the percent yield of the reaction? 4al + 3o₂ → 2al₂o₃ ?%

excess aluminum reacts with 3.50 moles oxygen to form aluminum oxide. the reaction generated 105 g al₂o₃. what is the percent yield of the reaction? 4al + 3o₂ → 2al₂o₃ ?%

Answer

Answer:

60.0%

Explanation:

Step1: Determine moles of $Al_2O_3$ from $O_2$

From the balanced equation $4Al + 3O_2\rightarrow2Al_2O_3$, the mole - ratio of $O_2$ to $Al_2O_3$ is $3:2$. Given $n_{O_2}=3.50$ mol, then $n_{Al_2O_3}=\frac{2}{3}\times n_{O_2}=\frac{2}{3}\times3.50$ mol $\approx2.33$ mol.

Step2: Calculate theoretical mass of $Al_2O_3$

The molar mass of $Al_2O_3$ is $M=(2\times26.98 + 3\times16.00)$ g/mol $= 101.96$ g/mol. The theoretical mass $m_{theo}=n_{Al_2O_3}\times M = 2.33$ mol $\times101.96$ g/mol $\approx237$ g.

Step3: Calculate percent - yield

The percent - yield formula is $\text{Percent Yield}=\frac{m_{actual}}{m_{theo}}\times100%$. Given $m_{actual} = 105$ g and $m_{theo}\approx237$ g, then $\text{Percent Yield}=\frac{105}{237}\times100%\approx60.0%$.