Theoretical and percent yield help

Determine the theoretical and percent yield of hydrogen gas if 75.0 g water undergoes electrolysis to produce hydrogen and oxygen and 3.43 g hydrogen is collected.

theoretical yield in grams

percent yeild in percent

Just for argument's sake let us assume that all the gases are collected at STP

Step 1: Balanced equation. 2 H2O -----------------> 2H2 + O2

Step 2: Convert masses to moles

mass of water = 75.0 g therefore moles of water = 75.0 g / 18.02 g/mol = 4.16 moles

Step 3: use the mole ratio between water and H2 from the balanced equation to calculate the expected moles of H2.

mole ratio is 1 to 1, therefore the moles of H2 should also be 4.16 moles

Step 4: Convert 4.16 moles of H2 into a mass

mass = moles x molar mass

mass = 4.16 moles x 2.02 g/mol = 8.41 grams. This is your theoretical yield

STep 5: Your actual yield is 3.43 g Therefore the percent yield is (8.41 - 3.43) x 100 / 8.41 = 59% yield

Theoretical yield is the amount of products you could obtain in case the reaction proceed comlpetely to the reagent formation according to its equation.

In this case: 2H2O --> 2H2 + O2

That means one 2 moles of water decomposes to 2 moles hydrogen and one mole oxygen

75 g water correspond to 75/18=4.17 moles water (18= molecular weight of water), that decompose to 8,34 noles H2 and 4.17 moles O2.
Theoretical yeld of Hydrogen in grams therefore is: 8.34 moles x 2 g/moles = 16.68 g

Percent yield is the ratio between obtained and theoretical amount of products. In this case, for hydrogen, 3.43 / 16.68 x100 = 20.56%

I hope it helps.

Regards.

Paolo