Sodium Carbonate crystals (27.82g) were dissolved in water and made up to 1dm^3. 25ml of the solution were neutralized by 48.8ml of HCl 0f 0.1M. Find n in the formulae of Na2CO3 crystals.

1 Answer

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`Na_2CO_3+2HCl rarr 2NaCl+CO_2+H_2O`


Amount of HCl consumed `= 0.1/1000xx48.8`

Amount of `Na_2CO_3` reacted `= 1/2xx0.1/1000xx48.8 = 0.00244`


So we have 0.0244mol in 25ml of `Na_2CO_3` solution.

Amount of `Na_2CO_3` in `1dm^3 = 0.00244xx40 = 0.0976`


Mass of 0.0976 `Na_CO_3` moles `= 106xx0.0976 = 10.3456g`


Mass of water in `Na_2CO_3` crystals `= 27.82-10.3456 = 17.4744g`


Amount of water moles `= 17.474/18 ~~ 1`


So the answer is n = 1