The electrolysis of water forms h2 and o2. 2h2o mc017-1.jpg 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%

1) Chemical equation

2H2O > 2H2 + O2

2) theoretical molar ratios:

2 mol H2O : 2 mol H2 : 1 mol O2

3) determine the number of moles in 17.0 g of H2O and 10.2 g of 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

4) theoretical yield

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

5) Percent yield

Percent yield = (actual moles / theoretical moles) * 100 = (0.319 / 0.472)*100 = 67.6%

the fourth option 67.6%

The reaction will shift to the right