[Kris44]

Ok i understand. I have just started studying moles etc so am still getting my head around it .

I have another question that i have attempted but am not sure if i'm correct.

What volume of hydrochloric acid of concentration 2.00mol dm³ would have to be added to 25.0cm³ of 0.500mol dm³ sodium carbonate to give a neutral solution of sodium chloride?

Heres what i have done.

I worked out equation to be Na₂CO₃ + 2HCL ---->2NaCl + CO₂+H₂O

So 1mole neutralises 2moles of acid?
HCL con =2.00mol dm₃
moles of sodium carbonate= 25.0x0.5/1000 = 1.25^-2
=1.25x10^-2 x 2 =0.025=2.5x10^-2
moles of acid =VM/1000
=V=moles x 1000/M = 12.5cm³??

This is probably completely wrong but i have had a go.

[Borek]

Perfect