Hi, I need some help.
In the example given for finding the percent yield of the 2 H20 > 2 H2 + O2 equation, it calculates the theoretical yield as 3.33g H2.
Then in the percent yield equation to follow, it replaces the "theoretical yield" with 1.67g H2 (half of 3.33g H2). Can anyone explain to me why this happens?
The book doesn't provide any detail and it is racking my mind!
Thanks in Advance!
Stoichiometry/Percent Yield (Chem p46)

 Posts: 47
 Joined: Wed May 27, 2020 7:25 pm
Re: Stoichiometry/Percent Yield (Chem p46)
Hi!
Sorry for any confusion, it's possible that you have a slightly older version of the books and there is a typo which is now corrected in the newer version. To go through the calculation, the example says that 1.31 g of hydrogen gas were produced from 30 g of water. To get theoretical yield: 30 g H2O x (1 mol H2O / 18 g) x (1 mol H2 / 2 mol H2O) x (2 g / 1 mol H2) = 1.67 g H2. This is then plugged in to get percent yield : 1.31/1.67 x 100% = 78.4%.
I hope this helps!
Sorry for any confusion, it's possible that you have a slightly older version of the books and there is a typo which is now corrected in the newer version. To go through the calculation, the example says that 1.31 g of hydrogen gas were produced from 30 g of water. To get theoretical yield: 30 g H2O x (1 mol H2O / 18 g) x (1 mol H2 / 2 mol H2O) x (2 g / 1 mol H2) = 1.67 g H2. This is then plugged in to get percent yield : 1.31/1.67 x 100% = 78.4%.
I hope this helps!
Re: Stoichiometry/Percent Yield (Chem p46)
I didn't expect such a speedy response! Lol
But the balance equation gives the mole ratio: 2 mol H2 to 2 mol H2O. So the theoretical yield will still calculate to what the book says (3.33 g H2) :/
Is there something I'm missing
But the balance equation gives the mole ratio: 2 mol H2 to 2 mol H2O. So the theoretical yield will still calculate to what the book says (3.33 g H2) :/
Is there something I'm missing

 Posts: 57
 Joined: Fri May 29, 2020 11:43 pm
Re: Stoichiometry/Percent Yield (Chem p46)
Hey! I've attached a screenshot of my textbook just to make sure that our versions are the same. If they are, then the 3.33 value is the theoretical yield, or the max amount of product that you will be able to get from this reaction. However, the question stem tells us that the actual yield is 1.31, which is the amount that they were actually able to get. Therefore, you divide the 1.31 value by 3.33 to find out the percent yield, or how much of the theoretical yield you were able to obtain. The book was right that there is a 2:2 (or 1:1) ratio between H2O and H2 in the balanced equation, but I think Nancy was also right in pointing out that the 1.67 value was probably a typo.
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