Stuck on Chemistry calculation

I would really appreciate it if someone could help me on the following question:

sodium carbonate crystals (27.8230g) were dissolved in water and made up to 1.00 dm3. 25.0 cm3 of the solution were neutralised by 48.8 cm3 of hydrochloric acid of conc 0.100 mol dm-3. Find n in the formula Na2CO3.nH2O

thanks guys

Reply 1

oxymoron

7

Work out how many moles of HCl were used.

One carbonate reacts with one acid ... so this is the moles of carbonate present in 25cm3.

Scale this up to the moles in 1dm3 and this is the number of moles in 27.82g.

From this you can work out the relative formula mass which will give you n.

If you still have any questions about an individual step then let me know.

Reply 2

bachelor

OP

you know when you scaling up do you times the moles in 25cm3 by 40 so that you can determine the moles in 1dm3?
i tried this and i dont get the answer required

Reply 3

oxymoron

7

What am I talking about?!

The eqation is

Na2CO3 + 2HCl --> NaCl + CO2 + H2O

So TWO moles of acid react with one mole of Na2CO3

Reply 4

bachelor

OP

oxymoron

What am I talking about?!

The eqation is

Na2CO3 + 2HCl --> NaCl + CO2 + H2O

So TWO moles of acid react with one mole of Na2CO3

how about the scaling factor.... i'm really stuck, have you tried working it out?

Reply 5

idiopathic

15

What's the answer?

Reply 6

bachelor

OP

endeavour

What's the answer?

Na2CO3.10H2O

n = 10
but i dont know how to get the answer?

Reply 7

charco

Study Forum Helper

18

bachelor

I would really appreciate it if someone could help me on the following question:

sodium carbonate crystals (27.8230g) were dissolved in water and made up to 1.00 dm3. 25.0 cm3 of the solution were neutralised by 48.8 cm3 of hydrochloric acid of conc 0.100 mol dm-3. Find n in the formula Na2CO3.nH2O

thanks guys

48.8 cm3 of 0.1M HCl = 0.00488moles

Na2CO3 + 2HCl --> NaCl + CO2 + H2O

therefore moles of Na2CO3 = 0.00488/2 = 0.00244moles

This is in 25cm3 therefore the moles in 1000cm3 = 0.00244/0.025 =0.0976moles

If the formula = Na2CO3.nH2O

Then the neutralisation has measured only the Na2CO3

Therefore the mass of Na2CO3 = RMM x no of moles = 106 x 0.0976 = 10.3456g

The remaining mass must be due to water = 27.823 - 10.3456 = 17.4774g

RMM of water = 18 therefore this is equivalent to 17.4774/18 moles = 0.971

Thuus the mole ratio of Na2CO3 to water in the original compound = 0.096 : 0.971

or approximately 1 :10

The formula is therefore Na2CO3.10H2O

Reply 8

bachelor

OP

thanks dude

Stuck on Chemistry calculation

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