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What volume of .0512 M Ca(OH)2

  1. Sep 22, 2005 #1
    Tough Problem

    I need help with this problem. Using the chemical equation
    HNO_2 + Ca(OH)2 = H20 + Ca(NO_2)2, What volume of .0512 M Ca(OH)2 is required to react with 0.2 g of HNO_2? Could you please also show me how to get the answer.
    Last edited: Sep 22, 2005
  2. jcsd
  3. Sep 22, 2005 #2


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    Please first show us what you've tried so far.

    What do you know about attacking problems like this ? What have you been taught and/or what does your text have to say about it ?
  4. Sep 25, 2005 #3
    i don't really know much at all thats why I need help
  5. Sep 25, 2005 #4


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    HNO_2 + Ca(OH)2 = H20 + Ca(NO_2)2

    First of all, one needs to write a correct equation, Ca(NO_2)2 has 2 (NO_2), so the left side require 2 HNO_2. BTW, is that HNO_2 (nitrous) or HNO_3 (nitric) acid?

    With this equation, one using moles. When converting mass (g) to moles, one uses the atomic or molecular mass, e.g. 1 mole of H2 = 2 grams, and 1 mole of H2O = 18 grams.

    from hyperphsics concepts - http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

    For some examples of chemical equations, see - http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/reactcon.html#c1
  6. Sep 25, 2005 #5


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    Well, you really need to learn at least an entire chapter then. Doing one problem for you will not get you terribly far. I suggest you study the chapter(s) in your chemistry text that deals with the mole concept and stoichiometry.

    Really, that's the ONLY way to learn the subject....besides paying attention to what happens in classes.
  7. Sep 26, 2005 #6
    Gokul43201 is correct (I do believe that reading books in general is one of the best way to learn-->or just in my opinion, textbooks can indeed teach better than teachers at times :rolleyes:)

    As Astronuc suggested, I think your equation should be:
    [tex] {\text{2HNO}}_{\text{2}} + {\text{Ca}}\left( {{\text{OH}}} \right)_2 \to 2{\text{H}}_{\text{2}} {\text{O}} + {\text{Ca}}\left( {{\text{NO}}_{\text{2}} } \right)_2 [/tex]

    If reaction goes to completion, then to find the volume of [itex] 0.0512M\;{\text{Ca}}\left( {{\text{OH}}} \right)_2 [/itex] needed to completely react with [itex] 0.2g\; {\text{HNO}}_{\text{2}} [/itex], you might want to:

    >>1) Convert [itex] 0.2g\; {\text{HNO}}_{\text{2}} [/itex] to moles of [itex] {\text{HNO}}_{\text{2}} [/itex].

    >>2) Find out from the balanced equation how many moles of [itex] {\text{Ca}}\left( {{\text{OH}}} \right)_2 [/itex] you will need to react with the [itex] {\text{HNO}}_{\text{2}} [/itex].

    >>3) Find out how many liters of [itex] 0.0512M\;{\text{Ca}}\left( {{\text{OH}}} \right)_2 [/itex] are needed to comprise that many moles of [itex] {\text{Ca}}\left( {{\text{OH}}} \right)_2 [/itex].

    >>Put it all together, and you'll have:

    [tex] \left( {\frac{{0.2g\,{\text{HNO}}_{\text{2}} }}{{\text{1}}}} \right)\left( {\frac{{1{\text{ mol HNO}}_{\text{2}} }}{{47g\;{\text{HNO}}_{\text{2}} }}} \right)\left( {\frac{{{\text{1 mol Ca}}\left( {{\text{OH}}} \right)_2 }}{{{\text{2 mol HNO}}_{\text{2}} }}} \right)\left( {\frac{{{\text{1}}\,{\text{L}}\;{\text{Ca}}\left( {{\text{OH}}} \right)_2 }}{{0.0512{\text{ mol}}\;{\text{Ca}}\left( {{\text{OH}}} \right)_2 }}} \right) = \boxed{0.04\,{\text{L}}\;{\text{of }}0.0512M\;{\text{Ca}}\left( {{\text{OH}}} \right)_2 } [/tex]

    And your answer will have one significant figure ---> due to those [itex] 0.2g\;{\text{HNO}}_{\text{2}} [/itex] :wink:

    Hope this helps! :smile: :shy:
    (shouldn't this be in the homework section tho? :redface:)
    Last edited: Sep 26, 2005
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