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

[danne89]

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:
$$\frac{x}{Na_2B_4O_7 * 10 H_2O} = \frac {31}{4 H_3BO_3}$$
x = 47.8 g

$$\frac {0.2x}{3HCl} = \frac {31}{4 H_3BO_3}$$
x = 275 g

 

[dextercioby]

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

Daniel.

 

[dextercioby]

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

Daniel.

 

[danne89]

I didn't write down the reaction formula. I just used the chemical equality-method.

 

[dextercioby]

What's that?

Daniel.

 

[Gokul43201]

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.

 

[dextercioby]

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

Daniel.