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Sodium Carbonate crystals (27.82g) were dissolved in water and made up to 1dm^3. 25ml...

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lak-86 | Student, Undergraduate | Salutatorian

Posted July 12, 2013 at 7:04 PM via web

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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.

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

Posted July 12, 2013 at 7:32 PM (Answer #1)

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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

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