Answer
b. 95%
Explanation
Given:
Mass of K₂O produced (actual yield) = 28.56 g
Mass of K that reacted = 25.00 g
Equation: 4K(s) + O₂(g) → 2K₂0(s)
What to find:
The percent yield of K₂O.
Step-by-step solution:
The first step is to calculate the theoretical yield of K₂O produced.
From the balanced equation, 4 mol K produced 2 mol K₂O
Molar mass of K₂O = 94.20 g/mol)
Molar mass of K = 39.10 g/mol)
This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O
So 25.00 g K will produce:
![\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O](https://tex.z-dn.net/?f=%5Cfrac%7B25.00%5Ctext%7B%20%7Dg%5Ctext%7B%20%7DK%5Ctimes188.40%5Ctext%7B%20%7Dg%5Ctext%7B%20%7DK%E2%82%82O%7D%7B156.40%5Ctext%7B%20%7Dg%5Ctext%7B%20%7DK%7D%3D30.1151%5Ctext%7B%20%7Dg%5Ctext%7B%20%7DK%E2%82%82O)
Actual yield of K₂O = 28.56 g
Theoretical yield of k₂O = 30.1151 g
The percent yield for the reaction can now be calculated using the formula below:
![\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20Percent%5Ctext%7B%20%7Dyield%3D%5Cfrac%7BActual%5Ctext%7B%20%7Dyield%7D%7BTheoretical%5Ctext%7B%20%7Dyield%7D%5Ctimes100%5C%25%20%5C%5C%20%20%5C%5C%20Percent%5Ctext%7B%20%7Dyield%3D%5Cfrac%7B28.56%5Ctext%7B%20%7Dg%7D%7B30.1151%5Ctext%7B%20%7Dg%7D%5Ctimes100%5C%25%20%5C%5C%20%20%5C%5C%20Percent%5Ctext%7B%20%7Dyield%3D0.9484%5Ctimes100%5C%25%20%5C%5C%20%20%5C%5C%20Percent%5Ctext%7B%20%7Dyield%3D94.84%5C%25%5Capprox95%5C%25%20%5Cend%7Bgathered%7D)
Therefore, the percent yield for the reaction is 95%.