Answer:
Percent Yield K₂O = 59.2%
Step-by-step explanation:
Before you can calculate the percent yield, you need to find the theoretical yield. This value is the amount of product produced using the balanced chemical equation and molar masses. However, the question does not specify which reactant is the limiting reagent. Therefore, you have to calculate the mass of the product starting from both values.
To find the theoretical yield, you need to (1) convert grams K/O₂ to moles K/O₂ (via molar masses), then (2) convert moles K/O₂ to moles K₂O (via mole-to-mole ratio from balanced equation coefficients), and then (3) convert moles K₂O to grams K₂O (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given values.
Molar Mass (K): 39.098 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
Molar Mass (K₂O): 2(39.098 g/mol) + 15.998 g/mol
Molar Mass (K₂O): 94.194 g/mol
The balanced equation:
4 K + O₂ -----> 2 K₂O
8.92 g K 1 mole 2 moles K₂O 94.194 g
-------------- x ----------------- x -------------------- x ---------------- = 10.7 g K₂O
39.098 g 4 moles K 1 mole
3.28 g O₂ 1 mole 2 moles K₂O 94.194 g
---------------- x --------------- x --------------------- x -------------- = 19.3 g K₂O
31.996 g 1 mole O₂ 1 mole
Because potassium produces the smaller amount of product, it is the limiting reagent. This means that all the potassium reactant is used up before the oxygen gas runs out. So, the theoretical yield of potassium oxide is 10.7 grams.
Now, you can use the theoretical yield and actual yield to determine the percent yield.
Actual Yield
Percent Yield = ------------------------------ x 100%
Theoretical Yield
6.36 g K₂O
Percent Yield = ------------------------ x 100%
10.7 g K₂O
Percent Yield = 59.2%