please help with chemistry questions

[ed_saifa]

Hey
Question:
Anhydrous sodium carbonate was used in an experiment to determine the concentration of an unknown hydrochloric acid solution. A sample of the carbonate was heated, cooled and weighed until constant weight was obtained.

(a)After being treated as described above, 1.377g of anhydrous sodium carbonate was dissolved in water and made up to 250cm ^3 in a volumetric flask. A 25.00ml portion of this solution was titrated with hydrochloric acid, using methyl orange as indicator. The titration required 22.65mL of the acid for complete neutralization.
(i) Determine the molar concentration of the sodium carbonate solution.
(ii) Determine the molar concentration of the unknown hydrochloric acid solution

(b) A second solution was prepared by dissolving  0.5g of crystalline sodium carbonate in water. This solution was then titrated with another portion of the hydrochloric acid solution. It was found that 30.45mL of the acid was required for complete neutralization.
(i) Calculate the apparent molar mass of the sodium carbonate used in the procedure
(ii) Determine the molecular formula of the crystals used in this procedure

Thank you!

[Collin Li]

a)

i.



ii.



Chemical equation:

[Toothpaste]

 








Find the moles of in the titration
given: and



Using the equation

[ed_saifa]

thanks!

[Collin Li]

Toothpick, you're right. I have fixed this consequential error. I realised this when doing part b.

Thanks!

[Collin Li]

b)

Realise that crystalline sodium carbonate refers to hydrated sodium carbonate (is of the form ).

i.



Since the stoichometric ratio between anhydrous sodium carbonate and hydrochloric acid should be equivalent to the reaction with the hydrated form (because the additional waters do not change the acid-base reaction), we can say:





ii.

Using (dry sodium carbonate + an unknown amount of water):





Therefore, the molecular formula is:

[ed_saifa]

Thanks heaps

[ed_saifa]

Hi, i've got another questions. Help would be hot! XD Thanks in advance
Question 5.
An aqueous solution of potassium dichromate, K2Cr2O7, was prepared by dissolving 2.31 g of the pure solid in enough water to make up 500 mL of solution.
A piece of iron wire, weighing 0.597 g was added to a dilute acid solution so that all the Fe(s) atoms were converted to Fe2+(aq) ions. The resulting solution was diluted to 100 mL.
25.0 mL aliquots of the Fe2+(aq) solution were further acidified and titrated with the potassium dichromate solution.
During the titration Fe2+(aq) ions were oxidised to Fe3+(aq) ions by the Cr2O72-(aq) ions, which were converted to Cr3+(aq) ions.
The average titre was 27.3 mL
(a)   Write a balanced equation for the conversion of Fe(s) to Fe2+(aq) in acidic solution
(b)   The titration involves a redox reaction. Write (I) the oxidation half-equation, (ii) the reduction half-equation, and (iii) the overall redox equation describing this reaction.
(c)   Determine the concentration of the potassium dichromate solution.
(d)   Determine the percentage purity of the iron wire.

[Collin Li]

Here's the first few parts (the theory parts)

For the redox equations with iron, just use commonsense (balance the atoms on both sides) and/or refer to the electrochemical series:

a)


b)

i) oxidation:

ii) This one we have to work out: it goes from to so we use those balancing rules for redox (balance oxygens with waters, then hydrogen atoms with , and charges with electrons). You should get:

reduction:

iii) Adding 6 times the oxidation reaction and 1 times the reduction reaction (to balance the number of electrons) yields:

overall reaction:

[Collin Li]

c)

d) Find the amount of potassium dichromate consumed in reaction with iron:





Recognise that this is only the number of moles of iron in one aliquot (25.0 mL):

[ed_saifa]

Thanks heaps Coblin!!

[ed_saifa]

Could anyone please help me with this question?

One molecule of vitamin B contains 63 carbon atoms. The percentage by mass of carbon in Vitamin B is 55.74 %. What is the molar mass of vitamin B?

Thanks

[midas_touch]

Mass of 63 carbon atoms: 63*12 = 756 amu.

Percentage of carbon in Vitamin B = 55.74%

Hence, molar mass of Vitamin B is given by
756/(0.5574) g/mol
= 1356.3 g/mol

Keeping in mind that 1 amu is exactly equal to 1 g/mol.

[ed_saifa]

thanks heaps midas_touch! i would raise your karma but i dont know how =(

[Toothpaste]

thanks heaps midas_touch! i would raise your karma but i dont know how =(

You need 50 posts to do that.