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Topic: balancing an equation  (Read 26335 times)

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sundrops

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Re:balancing an equation
« Reply #15 on: October 03, 2005, 11:38:53 PM »
WOW!
oh my goodness - I GET IT! :D
thank you very much. it makes sense now.

thank you for being so patient with me   :)

sundrops

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Re:balancing an equation
« Reply #16 on: October 04, 2005, 12:06:35 AM »
ok now i need to find the theoretical yield of salicylic acid so I've gone about it this way:

0.5g methyl salicylate / 152g/mol = o.oo33 mol methyl salicylate

1g NaOH / 40g/mol NaOH = 0.025 mol NaOH

because it is all a 1:1:1:1 ratio that means that there is 0.0033 mol of sodium salicylate  produced because it is the limiting reagent.

so taking that into account, I got to part 2.
2NaC7H5O3 + H2SO4 --> 2C7H6O3 + Na2SO4

0.0033 mol sodium salicylate / 2 mol sodium salicilate * 160g/mol sodiumsalicylate = 0.264 g sodium salicylate = 0.264 g salicylic acid 1:1 ratio

does that look about right?

Offline mike

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Re:balancing an equation
« Reply #17 on: October 04, 2005, 12:19:43 AM »
The first part is right: 0.0033 moles of sodium salicylate.

Next, how many grams of sodium salicylate is 0.0033 moles?

Now, new reaction (equation two where you are adding acid):

1. how many grams of sodium salicylate? (from part one)
2. how many moles of sodium salicylate? (grams --> moles)
3. ratio of sodium salicylate to salicylic acid is? (from your second equation)
4. theoretical mass of salicylic acid (moles --> grams)
There is no science without fancy, and no art without facts.

sundrops

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Re:balancing an equation
« Reply #18 on: October 04, 2005, 12:36:16 AM »
thanks mike - you've been so helpful.
I feel like I understand whats happening now and I'm less stressed too  ;)

Offline mike

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Re:balancing an equation
« Reply #19 on: October 04, 2005, 12:46:16 AM »
No worries, glad to be of some help  :)
There is no science without fancy, and no art without facts.

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