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Hydrochloric acid solution is needed to make 31 g boric acid?

  1. Feb 14, 2005 #1
    Hi. I've problem with this question:
    How much Na2B4O7 * 10H2O and 20% hydrochloric acid solution is needed to make 31 g boric acid?

    My work:
    [tex]\frac{x}{Na_2B_4O_7 * 10 H_2O} = \frac {31}{4 H_3BO_3}[/tex]
    x = 47.8 g

    [tex]\frac {0.2x}{3HCl} = \frac {31}{4 H_3BO_3}[/tex]
    x = 275 g
     
  2. jcsd
  3. Feb 14, 2005 #2

    dextercioby

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    Are u sure that u've written the reaction correctly...?It should be:
    [tex] Na_{2}B_{4}O_{7}\cdot 10H_{2}O+2HCl \rightarrow 4H_{3}BO_{3}+2NaCl+5H_{2}O[/tex]

    Daniel.
     
  4. Feb 14, 2005 #3

    dextercioby

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    IIRC,there shouldn't be any boron chloride...

    Daniel.
     
  5. Feb 14, 2005 #4
    I didn't write down the reaction formula. I just used the chemical equality-method.
     
  6. Feb 14, 2005 #5

    dextercioby

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    What's that?

    Daniel.
     
  7. Feb 14, 2005 #6

    Gokul43201

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    What boron chloride ?

    Danne, you are not considering the stoichiometry of the problem. You can not solve it without writing down the balanced equation (like Dexter has done for you), and using the fact that 1 mole of sodium borate (or whatever it's called) gives 4 moles of boric acid.
     
  8. Feb 14, 2005 #7

    dextercioby

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    Natrium perborate,a.k.a. BORAX...

    Daniel.
     
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