titration!! HELP IMMEDIATLY NEEDED!!!!!!!?

Question

heres my equation: Na2CO3 + H2SO4 ---> Na2SO4 + H2O + CO2...is that right? now how do i find the amount of sodium carbonate i will be using? i can work out concentration later....at the moment im writing my plan for a titration i will be carrying out at school. i am doing AS level chemistry. PLEASE HELP! ive been up for hours already and i dont know how to do it!

mole ratio is 1:1 yeah? then wat? do i have to find the RFM of Na2CO3? or what??

HELP!!!!!!!!!!!!!!
thanks, need advice tonight, its in for tomorrow!
charl x

but how do you work out the amount of substance (sodium carbonate) that you will be using????

Answer 1

Try to think of things in a general sense first. Then, proceed to reason to specifics. First Relax and be calm.
(1st
Remember: an acid plus a base will always yield a salt plus water.
(2nd
Remember:
Whenever you have three pieces of information you can always find the third through quantitative analysis and multiplying across the equal sign. Use a simple example:

First balance the equation in terms of the number of moles of each you will need. Let sodium carbonate eual "X". Solve your equation and make sure it is balanced.

Answer 2

The equation is correct.
The mole ratios are all 1:1 :: 1:1:1
Find RFM (Na2CO3)
2 x Na = 2 x 23 = 46
1 x C = 1 x 12 = 12
3 x O = 3 x 16 = 48
46 + 12 + 48 = 106

For a titration you need to know the masses of the reactants. Are you are trying to determine the concentration of H2SO4?

Weigh out 2.12 grams of Na2CO3
Moles(Na2CO3) is 2.12 divided by 106 = 0.02

This 0.02 moles of Na2CO3 is dissolved into 250 cm^3 of distilled water, in a volumetric flask.
Conc'n (Na2CO3) = 0.08 M (Mol dm^-3)

A 25 cm^3 aliquot is taken from the volumetric flask and placed in a conical flask, together with a suitable indicator.

The moles(Na2CO3) in the 25 cm^3 aliquot is 0.002.

As the mole ratios are 1:1 then 0.002 moles (H2SO4) is required to neutralise the system, this is indicated by a colour change in the 'indicator'.

If, for the sake of argument, 40 cm^3 of H2SO4 are titrated off from the burette into the conical flask, then this 40 cm^3 contains 0.002 mol(H2SO4).

So Conc'n(H2SO4) = (0.002 x 1000)/40 = 0.05 mol dm^-3(H2SO4). Sometimes written as 0.05 M.

Hope this helps.!!!!

Answer 3

divide the two equations and then add 4 and then you have to try to find out the rfm of na2c03, i passed as level chemistry and am now opening up my own chemist