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Question 6B.11 (7th Edition)

Posted: Fri Jan 18, 2019 3:58 pm
by Lexie Baughman 2C
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!

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

Posted: Fri Jan 18, 2019 11:21 pm
by Matthew Tran 1H
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-.