Volumetric analysis Please help (:

When a 0.100g sample of a urea-based fertiliser was boiled with an excess of sodium hydroxide, the following reaction took place: NH2CONH2 + 2OH- -------> 2NH3 + CO3 (2-)

The ammonia gas evolved was absorbed in water, and the resulting aqueous solution titrated with hydrochloric acid. 14.00cm^3 of 0.200 mol/dm^3 of HCl were required for neutralisation.
NH3 + HCI ----> NH4Cl
What is the percentage of nitrogen in this sample of fertiliser?

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8.58g of washing soda crystals of formula, Na2CO3.xH2O, were made up to 250 cm^3 of aqueous solution. 25.0 cm^3 of this solution required 30.00cm^3 of 0.200 mol/dm^3 of HCl for neutralisation with methyl orange as indicator.
Calculate the value of X in the formula of washing soda.

Please help me explain in detail for each step taken! Thanks alot!!! Really appreciate the help !:D

NH3 + HCI ----> NH4Cl

moles of HCl = 0.200 mol/dm^3 x 14 / 1000 = 0.0028 moles

these are equal to the moles of NH3 which all came from the urea
mass of NH3 = 0.0028moles x 17 g/ mole = 0.0476 g
mass of N = 0.0476 x 14/ 17 = 0.0392g
%N = 0.0392 / .1 = 39.2 % N

2 HCl + Na2CO3 = 2 NaCl + H2O + CO2

moles of HCl = .2 mole liter^-1 x .030 liter = .006 moles

these react with = .006 / 2 moles of Na2CO3 = .003 moles
these are in 25.0 mL so moles in the flask = .003 * 250 / 25 = .03 moles
molar mass of Na2CO3 = 106 g / mole so mass of the salt that was put into the flask =
= .03 x 106 = 3.18g of Na2CO3

mass of hydrated salt = 8.58 so mass of water = 8.58 - 3.18 = 5.40 g or
5.40 / 18 moles = .3 moles of h2O

we have .03 moles of Na2CO3 and .3 moles of H2O

formula is Na2CO3 * 10 H2O