A 26.3 gram sample of solid potassium chlorate decomposed and produced 9.45 grams of oxygen gas. What is the percent yield of oxygen?
2KClO3 ==> 2KCl + 3O2
mols KClO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2 produced.
Now convert mols O2 to grams O2. g O2 = mols O2 x molar mass O2 = ? This is the theoretical yield (TY).
The actual yield (AY) is given as 9.45 g
%yield = (AY/TY)*100 = ?
To calculate the percent yield of oxygen, we need to compare the actual yield of oxygen with the theoretical yield.
Step 1: Find the theoretical yield of oxygen.
The molar mass of potassium chlorate (KClO3) is:
- Potassium (K): 39.1 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16 g/mol (1 oxygen atom)
So, the molar mass of potassium chlorate is:
(39.1 g/mol) + (35.45 g/mol) + (3 × 16 g/mol) = 122.55 g/mol
From the balanced chemical equation:
2KClO3 → 2KCl + 3O2
We can see that 2 moles of potassium chlorate produce 3 moles of oxygen gas. Therefore, the molar ratio is:
2 moles KClO3 : 3 moles O2
For the given mass of potassium chlorate (26.3 g), we need to convert it to moles using the molar mass:
26.3 g ÷ 122.55 g/mol = 0.214 moles KClO3
Using the molar ratio, we can find the theoretical yield of oxygen:
0.214 moles KClO3 × (3 moles O2 / 2 moles KClO3) × (32 g O2 / 1 mol O2) = 10.9 grams O2
Step 2: Calculate the percent yield of oxygen.
To find the percent yield, we divide the actual yield by the theoretical yield and multiply by 100:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
Given the actual yield of oxygen is 9.45 grams, we can now calculate the percent yield:
Percent Yield = (9.45 g / 10.9 g) × 100 = 86.7%
Therefore, the percent yield of oxygen is approximately 86.7%.
To calculate the percent yield of oxygen in this reaction, you need to compare the actual yield (9.45 grams) to the theoretical yield, which can be calculated based on the balanced chemical equation for the decomposition of potassium chlorate (KClO3):
2KClO3(s) → 2KCl(s) + 3O2(g)
From the equation, you can see that for every 2 moles of potassium chlorate decomposed, 3 moles of oxygen gas are produced. To find the theoretical yield of oxygen, you need to convert the given mass of potassium chlorate (26.3 grams) to moles and then use the mole ratio to find the corresponding moles of oxygen.
Let's calculate the moles of potassium chlorate:
Molar mass of KClO3 = atomic mass of K + atomic mass of Cl + 3 * atomic mass of O
= (39.1 g/mol) + (35.5 g/mol) + 3 * (16.0 g/mol)
= 122.5 g/mol
Moles of KClO3 = mass / molar mass
= 26.3 g / 122.5 g/mol
≈ 0.2142 mol
Using the mole ratio from the balanced equation, you can determine the moles of oxygen produced:
Moles of O2 = 3 * moles of KClO3
= 3 * 0.2142 mol
≈ 0.6426 mol
Now that you have calculated the theoretical yield of oxygen (0.6426 mol), you can calculate the percent yield using the formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Plugging in the values:
Percent yield = (9.45 g / (0.6426 mol * (32.0 g/mol))) * 100
≈ 46.4%
Therefore, the percent yield of oxygen in this reaction is approximately 46.4%.