The Percentage Yield of O2 in the Electrolysis of Water
The electrolysis of water forms H2 and O2
2H2O ➜ 2H2 + O2
What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
Use mc017-2.jpg.
15.1%
33.8%
60.1%
67.6%
- Chemical equation
- theoretical molar ratios:
- determine the number of moles in 17.0 g of H2O and 10.2 g of O2
- theoretical yield
- Percent yield
2H2O ➜ 2H2 + O2
2 mol H2O : 2 mol H2 : 1 mol O2
number of moles = mass in grams / molar mass
molar mass of H2O = 18.02 g/mol
molar mass of O2 = 32.00 g/mol
number of moles of H2O = 17.0 g / 18.02 g/mol = 0.943 moles
number of moles of O2 = 10.2 g / 32.0 g = 0.319 mol O2
2 mol H2O / 1 mol O2 = 0.943 mol H2O / x ➜ x = 0.943 mol H2O * 1 mol O2 / 2 mol H2O = 0.472 mol O2
Percent yield = (actual moles / theoretical moles) * 100 = (0.319 / 0.472)*100 = 67.6%
Answer: the fourth option 67.6%
What is the percent yield of O2 if 10.2 g of O2 is produced from the decomposition of 17.0 g of H2O?
The answer is D. 67.6%