25cm cube of 3grams of hydrated sodiumtrioxocarbonate(IV) .X H2O per 250cm cube was neutralised by 13.20cmcube of 0.16mol/dmcube HCl solution.write the equation of the reaction, calculate the concentration of base in mol/dmcube,calculate the molar mass of the base, find the value of X,find the percentage by mass of water of crystallization.
Na2CO3.xH2O + 2HCl ==> 2NaCl + xH2O + CO2 + H2O
millimols HCl = M x mL = 0.16 M x 13.20 mL = 2.112
1 mols Na2CO3.xH2O = 2 mols HCl; therefore, mols Na2CO3xH2O = 1/2 mols HCl and 1/2*2.11 = 1.056 millimols in the 25 cc solution taken for titration. The M of the base in that volume is M = millimols/mL = 1.056/25 = 0.0422 M. Another way to look at it is that you used just 0.1 (that's 25/250) so there are 10.56 millimols in the original solution so 10.56/250 = 0.0422 M for the base. So the 3 grams represents 10.56 millmols or 0.01056 mols. mols = grams/molar mass so 0.01056 mols = 3 g/molar mass and molar mass = 3/0.01056 = 284. molar mass Na2CO3 is approx 106 so water is 284-106 = 178 which represents 178/18 = 10 mols H2O so x in the formula is 10 and the formula is Na2CO3.10H2O
%H2O = (178/284)*100 = ?
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