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

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Discussion Overview

The discussion revolves around the calculation of the amounts of sodium borate (Na2B4O7 * 10H2O) and hydrochloric acid (HCl) solution required to produce 31 grams of boric acid (H3BO3). The focus is on the stoichiometry of the chemical reaction involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates that 47.8 g of sodium borate and 275 g of 20% hydrochloric acid solution are needed to produce 31 g of boric acid.
  • Another participant questions the correctness of the reaction equation used, suggesting it should be Na2B4O7·10H2O + 2HCl → 4H3BO3 + 2NaCl + 5H2O.
  • There is a mention that boron chloride should not be present in the reaction, indicating a potential misunderstanding of the products formed.
  • One participant emphasizes the importance of writing down the balanced equation and considering stoichiometry to solve the problem correctly.
  • There is a reference to "Natrium perborate," also known as borax, which may imply a different perspective on the reactants involved.

Areas of Agreement / Disagreement

Participants express differing views on the correct reaction equation and the stoichiometric calculations. There is no consensus on the approach to solving the problem, and multiple competing views remain regarding the chemical processes involved.

Contextual Notes

Some participants note the necessity of a balanced equation for accurate stoichiometric calculations, while others challenge the initial assumptions and calculations presented. The discussion reflects uncertainty about the correct reactants and products.

danne89
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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
 
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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.
 
IIRC,there shouldn't be any boron chloride...

Daniel.
 
I didn't write down the reaction formula. I just used the chemical equality-method.
 
dextercioby said:
IIRC,there shouldn't be any boron chloride...

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

Daniel.
 

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