Answer the question(s) below about the following reaction.
2H2O2 (l) → 2H2O(g) + O2(g)
If 50.0 grams hydrogen peroxide produce 15.0 g oxygen, what is the percent yield for this reaction?
To find the percent yield for this reaction, follow these steps:
Step 1: Calculate the theoretical yield of oxygen.
The molar ratio between hydrogen peroxide (H2O2) and oxygen (O2) in the balanced equation is 2:1. This means that for every 2 moles of H2O2 reacting, 1 mole of O2 is produced.
To calculate the number of moles of H2O2:
molar mass of H2O2 = 2(atomic mass of hydrogen) + 2(atomic mass of oxygen) = 2(1.008 g/mol) + 2(16.00 g/mol) = 34.016 g/mol
moles of H2O2 = mass of H2O2 / molar mass of H2O2 = 50.0 g / 34.016 g/mol
To calculate the moles of O2:
moles of O2 = 1/2 * moles of H2O2
To calculate the theoretical yield of O2:
mass of O2 = moles of O2 * molar mass of O2
Step 2: Calculate the percent yield.
The percent yield can be calculated by dividing the actual yield (given as 15.0 g O2) by the theoretical yield (calculated in Step 1) and multiplying by 100.
percent yield = (actual yield / theoretical yield) * 100
Now you can substitute the values and solve for the percent yield.
To find the percent yield for a reaction, you need to compare the actual yield to the theoretical yield. Theoretical yield is the maximum amount of product that could be obtained from the reactants according to the balanced chemical equation.
1. Calculate the molar mass of hydrogen peroxide (H2O2):
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Multiply the molar mass of hydrogen by 2 and the molar mass of oxygen by 2 to get the molar mass of H2O2:
(2 × 1.01 g/mol) + (2 × 16.00 g/mol) = 34.02 g/mol.
2. Convert the given mass of hydrogen peroxide to moles:
- Divide the given mass (50.0 g) by the molar mass calculated in step 1:
50.0 g / 34.02 g/mol = 1.47 mol.
3. Use the balanced chemical equation to determine the stoichiometry between hydrogen peroxide and oxygen. From the equation 2H2O2 (l) → 2H2O(g) + O2(g), we see that for every 2 moles of H2O2, we should get 1 mole of O2.
4. Calculate the theoretical yield of oxygen:
- Since the stoichiometry ratio is 2:1, we divide the moles of H2O2 by 2 to get the moles of O2:
1.47 mol H2O2 / 2 = 0.735 mol O2.
- Multiply the moles of O2 by the molar mass of oxygen to get the theoretical yield in grams:
0.735 mol O2 × 32.00 g/mol = 23.52 g O2.
5. Calculate the percent yield:
- Divide the actual yield (15.0 g) by the theoretical yield (23.52 g) and multiply by 100 to get the percent yield:
(15.0 g / 23.52 g) × 100 = 63.84%.
Therefore, the percent yield for this reaction is approximately 63.84%.
look at the theoritical
molesO2=1/2 (massH2O2)/molmassH2O2)
massO2=molesO2*32
percent yeild=15/massO2above