Question

Heres a good one ; ) :
A solution of sodium carbonate is made using 79.5 grams of sodium carbonate in 250 cm3 of solution. 36.0 cm3 of this solution will neutralise 42.0 cm3 of a solution of hydrochloric acid.

Calculate the molarity of the hydrochloric acid solution.

Answer 1

Firs i found the amount of moles of sodium carbonate in the 79.5 grams of it and it gave me 0.76 moles aprox.
Then i found the molarity of it in the 250 cm3 solution and it gave me 0.0304mol/dm3(liter)
Then i got the amount of moles in the 36.0 grams of sodium carbonate and it gave me 0.11 moles
So then i said: 0.11 moles of sodium carbonate will neutralise 42 cm3 of hydrochloric acid.
After that i don't know how to finish it

Answer 2

good one... :)

Answer 3

Anyone? : )

Answer 4

molarity? that's like molar mass right? or is this n?

Answer 5

Molarity is equal to moles of solute over dm3(or liters) of solution.
M=moles of solute (g)/ decimeter cubed of solution (dm3)

Answer 6

i probably just failed my chem exam them....whooops

Answer 7

lol its ok

Answer 8

M=mol/L !
@NotTim ya you probably would lol jk jk don't kill me :D

Answer 9

ok so you have sodium carbonate which is:
Na2CO3 right?
and you neutralize it with HCl right?
so equation goes as follows:
Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
so if you calculated amount of Na2CO3 in moles now you see the ratio and write following:
n(HCl)/n(Na2CO3) = 2/1 --> n(HCl) = 2 * n(Na2CO3)
and i think you got it from here, if not just say and ill continue ;)

Answer 10

It was so simple. Thank you so much