Molar fraction problem? (Help)

[Jennifer_88]

molar fraction problem?? (Help)

hi,

I need help with this problem, ( I've spent fairly enough time trying to figure it out but :(

anyways, I'm

given) the mass of Naoh and hcl
the fraction number of the two before reaction
find) the fraction number after reaction is completed

so the equation i believe should be Hcl+Naoh-----> NaCl+H20

i've all the information for before the reaction, i just do not know how to figure out the masses of the components after the reaction. in other words i do not know how to relate after reaction to the molar fraction before


thanks in advance

 

[symbolipoint]

Your problem description is not accurate. What kind of fraction do you mean in your given information? Exactly what numbers for each material is given, and which number of which material are you asked to find?

 

[symbolipoint]

Maybe this will help:

NaOH + HCl -------> NaCl + H₂O

THAT is the balanced reaction. You may need to compute the formula weights of NaOH, HCl, and maybe NaCl.

 

[Jennifer_88]

ok thanks for the replay

i've .1 kg hcl
.2 kg Naoh
molar fraction of hcl is .1
molar fraction of naoh is .25

and i want to find the molar number for the reaction which should be h2o and Nacl

please let me know if I'm clear enough

thanks :)

 

[Jennifer_88]

it's been along time since i did any chemistry and I'm facing this now.

i think i need to find a relation between the molar fraction before the reaction and the mass of Nacl and h2o. maybe there is a ratio or something

 

[symbolipoint]

ok thanks for the replay

i've .1 kg hcl
.2 kg Naoh
molar fraction of hcl is .1
molar fraction of naoh is .25

and i want to find the molar number for the reaction which should be h2o and Nacl

please let me know if I'm clear enough

thanks :)

The masses of the reactants do not correspond to the molar fractions. Sum of 0.1 and 0.25 is 0.35 for your given molar fractions. What is the WHOLE on which the given molar fractions are based? If the given masses of HCl and of NaOH are to be used, then your calculations must be based on those and the values given as "molar fractions" are unreliable. The exact problem description is necessary for readers to understand what is given and what is asked.

 

[Jennifer_88]

ok here is the problem description

one tenth of a kg of an aqueous solution of HCL is poured into .2 kg of an aqueous solution of NaOH solution. the mole fraction of the HCL solution was .1, whereas that of the NaOH solution was .25. what are the mole fractions of each of the components in the solution after the chemical reaction has come to completion?

I really appreciate your help :)

 

[Borek]

the mole fraction of the HCL solution was .1

This solution contains 1 mole of HCl per each 9 moles of water. Can you calculate amount of water and amount of HCl from this information and from the known mass of the solution? Do the same for the other solution, then it will be just a stoichiometry problem, and calculating molar fractions would be a simple final touch.

 

[Jennifer_88]

i just found that it is actually not working.

i tried really different things to make it work but it has not

so what i did is

i did a ratio between the relationship you gave me 1 mole of hcl/ 9 mole of h2o = [(the mass of hcl (100 g?)/ (atomic mass36.46)]/[(mass of H2O/(atomic mass of h2o (18)]

then found the mass of h2o

then found the number of moles

then did the same thing for the NaOH to find the mass of Nacl

then did the mole fraction but I'm not getting the right answer :(

 

[Borek]

It must work, you are just missing something - you wrote you found mass of H₂O - but you can't find it from one equation.

$$\frac 1 {10} = \frac {n_{HCl}} {n_{HCl} + n_{H_2O}}$$

and

$$36.5n_{HCl} + 18n_{H_2O} = 100g$$

that should give you moles of both HCl and H₂O in the HCl solution.

 

[Jennifer_88]

well thank you for the help, but it just not working the final answer of molar fraction of H2o is .84 but I'm not getting that answer.

 

[symbolipoint]

well thank you for the help, but it just not working the final answer of molar fraction of H2o is .84 but I'm not getting that answer.

Look again carefully at what is explained in post #8. That should help you make progress. Note cafefully, "mole fraction", and not "mole ratio".

Borek - in your later post, did you really mean $$\frac{1}{10}$$ instead of $$\frac{1}{9}$$ ?

 

[Borek]

Borek - in your later post, did you really mean $$\frac{1}{10}$$ instead of $$\frac{1}{9}$$ ?

Thanks, my mistake. Corrected.

 

[symbolipoint]

Very good, Thanks Borek...

Jennifer_88,
The problem description is much better, perfectly understandable.

Compute the molecular or formula weights for HCl, NaOH, and H₂O. Your results should
be 36.46, 39.99, and 18.015, in that order.

Use your solutions' mole fractions to determine how much by moles of each solute
in gram of solution (each solution before mixing).

$$\[ \begin{array}{l} {\rm mole fraction reactant} = \frac{{moles\,solute}}{{moles\,solute\, + \,moles\,solvent}} \\ {\rm moles fraction solvent = }\frac{{1 - moleFraction{\mathop{\rm Re}\nolimits} ac\tan t}}{{moles\,solute + moles\,solvent}} \\ \end{array} \]$$

In the denominators of each of those, you use molecular weights; you use $$\[ {\rm mass of compound = }(molesOfCompound) \cdot (molecularWeightCompound) \]$$ giving you grams for each term in denominator.

You should now have been able to find the moles each solute and of solvent per
gram of solution FOR EACH solution BEFORE mixing and reacting. Moles of compound
per gram of solution.

Next you examine what happens in the reaction when the 0.1 kg of HCl solution
and the 0.2 kg of the NaOH solution are mixed together. HOW MANY MOLES of HCl,
and how many moles of NaOH? Continue by accounting for what has reacted, and
what if any if not reacted (because of being in excess). You now account for
all of the moles of compounds present: Salt, Water, and any unreacted compound.

Can you fill in and finish the rest of the process? I also used a table
of information to summarize the solution process as I went.
You should find mole fraction values very near, to two significant figures, of 0.0040 for
the NaCl, 0.13 for the unreacted compound (you figure out which this is),
and 0.87 for water.

 

[Borek]

$$moles fraction solvent = \frac{{1 - moleFractionReactant}}{{moles\,solute + moles\,solvent}}$$

I think you overdid with this one. Either just 1-x (when there is nothing else in the solution but substance and solvent), or

$$moles fraction solvent = \frac{moles_{solvent}}{total moles}$$

 

[symbolipoint]

I think you overdid with this one. Either just 1-x (when there is nothing else in the solution but substance and solvent), or

$$moles fraction solvent = \frac{moles_{solvent}}{total moles}$$

Yah, I tried to be thorough about some parts, but could only comfortably be descriptive about other parts. The problem-solving was easier on paper. On paper was much more comfortable, not so many buttons and code to select through a machine.

 

[Jennifer_88]

well i am going to work on it again right now and hopefully get .84 as an answer. wish me luck thanks guys you've been so helpful