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.

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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.

Chemistry
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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
good one... :)
Anyone? : )

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molarity? that's like molar mass right? or is this n?
Molarity is equal to moles of solute over dm3(or liters) of solution. M=moles of solute (g)/ decimeter cubed of solution (dm3)
i probably just failed my chem exam them....whooops
lol its ok
M=mol/L ! @NotTim ya you probably would lol jk jk don't kill me :D
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 ;)
It was so simple. Thank you so much

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