[jray]

Good day everybody.
First off all, i'm not very good at chemistry and i don't know how to solve this problem:
How many litres of 10% HCL solution (d=1.07g/cm3) it is possible to neutralise with 16 kg of NaOH?
I know it looks easy for most of you, but i couldn't solve it.

[AWK]

Start from a balanced equation of neutralization reaction

[jray]

HCl(aq) + NaOH -> H2O(l) + NaCl(aq)

[Arkcon]

Well done, now you a balanced equation.  The units used in chemistry for a balanced equation is moles.  You now have to convert % HCl and kg of NaOH to moles.  It's just like converting imperial units to metric. (Hint: why'd they give you density of HCl? What are it's units, and how do they relate to moles?)

[jray]

M(NaOH) = 16000 x 1mol/40 = 400 moles

10g HCl solute - 100g solution
x g - 1070 g     
-----
x=107g/l

M(HCl)=107/36=2.97mol/l

Correct me if i'm wrong

[Borek]

M(NaOH) = 16000 x 1mol/40 = 400 moles

Thats OK.

10g HCl solute - 100g solution
x g - 1070 g     
-----
x=107g/l

M(HCl)=107/36=2.97mol/l

Correct me if i'm wrong

Hard to say - you have correctly calculated amount of HCl in 1L of solution - but that's not what the questions asks for, although it can be a convenient intermediary result.

[jray]

So what i need to do next?

[AWK]

Stoichiometry of reaction says you 1 mole HCl react with 1 mole of NaOH. Hence you need 400 moles of HCl.
Now you can calculate volume of HCl. (2.97 mole in 1 liter)

[jray]

Stoichiometry of reaction says you 1 mole HCl react with 1 mole of NaOH. Hence you need 400 moles of HCl.
Now you can calculate volume of HCl. (2.97 mole in 1 liter)

So just a simple division (400/2.97) or I misunderstood something?

[AWK]

Just a simple division!

400 [mol] / 2,97 [mol/L] = xxx [L]

[jray]

So this is final answer? Hope it is ok, thank you for the help then