Chemistry Chapter 3 pg 46

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mikapaprika
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Joined: Sun Jan 19, 2020 12:28 am

Chemistry Chapter 3 pg 46

Post by mikapaprika » Sun Jan 19, 2020 12:33 am

Hi!

My question is in regards to the example we go through on page 46 using percent yield. I am having trouble understanding why we must divide 3.33g of H2 by 2 to get 1.67g H2 before entering it as the theoretical mass in the equation.
If we aren't determining the mass of one hydrogen to input into the equation, why is it being divided by 2? Why isn't it just 3.33g?

Thank you for your help! They skipped a step of explanation in the book so it was confusing to me.

Mika ;)
NS_Tutor_Mathias
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Re: Chemistry Chapter 3 pg 46

Post by NS_Tutor_Mathias » Mon Jan 20, 2020 4:43 pm

You didn't miss anything. An error got fixed here in the preceding equation between the previous edition of the books and this one, but apparently the follow-up equation slipped by unnoticed. This should be 1.31g/3.33g as you would assume!

(Previous editions had a much sneakier error that assumed only 1 mol H2 was made per 2 mol H2O, which is not true)
jeffbassett87
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Joined: Mon Jan 13, 2020 7:38 pm

Re: Chemistry Chapter 3 pg 46

Post by jeffbassett87 » Mon Jan 20, 2020 7:41 pm

I just read this today and was wondering the exact same thing.
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

Re: Chemistry Chapter 3 pg 46

Post by NS_Tutor_Mathias » Mon Jan 20, 2020 9:20 pm

jeffbassett87 wrote:
Mon Jan 20, 2020 7:41 pm
I just read this today and was wondering the exact same thing.
Funny enough, nobody noticed it before the first equation got fixed. Now it pops out!
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