Question 6B.11 (7th Edition)

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Lexie Baughman 2C
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Joined: Fri Oct 05, 2018 11:16 pm

Question 6B.11 (7th Edition)

Postby Lexie Baughman 2C » Fri Jan 18, 2019 3:58 pm

My question is in regards to Part B, where it asks the mass of solid Na2O added to the original flask. I'm not understanding why you need to multiply by the molar ratio, 1 mol Na2O/2 mol NaOH, if we're just finding the mass of Na2O. Thanks for the help!

Matthew Tran 1H
Posts: 124
Joined: Thu Sep 27, 2018 11:16 pm

Re: Question 6B.11 (7th Edition)

Postby Matthew Tran 1H » Fri Jan 18, 2019 11:21 pm

It'll make more sense if you write out the balanced chemical reaction: Na2O(s) + H2O(l) --> 2NaOH(aq). From this equation you can see that 1 mole of Na2O produces 2 moles of NaOH(aq), which is code for 2 moles of Na+ and OH-. Therefore the moles of Na2O is half that of OH-.


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