anonymous
  • anonymous
If 65.0 grams of Ca(OH)2 react with an excess of NaBr, how much NaOH is produced?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Here is some info you may need, NaBr+Ca(OH)2----------------->CaBr2+NaOH
anonymous
  • anonymous
first find moles of Ca(OH)2 in 0.65 grams=0.65/(40+(2(16+1)) by stoichiometry moles of Ca(OH)2=moles of NaOH mass of NaOH= Moles x Mr
anonymous
  • anonymous
get it??

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anonymous
  • anonymous
what's Mr?
anonymous
  • anonymous
Mr=molar mass=mass of 1 mole of a substance for NaOH Mr=23+16+1 =40
anonymous
  • anonymous
how do I find moles of Ca(OH)2 in 0.65 grams?
anonymous
  • anonymous
moles=mass/Mr =0.65/(74) where 74 is Mr=40+2(16+1) see my first post
anonymous
  • anonymous
ok but how do I find out how much NaOH is produced? still alil confused.....................
anonymous
  • anonymous
you have moles of Ca(OH)2?
anonymous
  • anonymous
i am confused
anonymous
  • anonymous
can we break it down again from the beginning?
anonymous
  • anonymous
Ok Step 1 Find Mr of Ca(OH)2
anonymous
  • anonymous
Step 2 Find Moles of Ca(OH)2
anonymous
  • anonymous
Step 3 From equation Moles of NaOH=Moles of Ca(OH)2 (1:1 in the equation)
anonymous
  • anonymous
Step 4 Mass of NaOH=Moles (found in previous step) x Mr of NaOH
anonymous
  • anonymous
no it's actually 74.09= Mr of Ca(OH)2
anonymous
  • anonymous
no Mr of Ca(OH)2 =mass of Ca + 2(mass of O + mass of H) =40+2(16+1) =74.09
anonymous
  • anonymous
follow these steps and see if you can achieve the answer
anonymous
  • anonymous
Ok so for step 2, in order to find the moles I need to divide .65/74?
anonymous
  • anonymous
I get 0.878
anonymous
  • anonymous
that's right
anonymous
  • anonymous
and I multiply that by 39.997
anonymous
  • anonymous
I get 35.1325
anonymous
  • anonymous
yes and you will get your answer
anonymous
  • anonymous
so is my final answer 35.1325 grams of NaOH?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ty!!!!!!!!!!!!!!!!!!!!!
anonymous
  • anonymous
:)
anonymous
  • anonymous
:)

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