hard titration alevel chem Q!

hello, im really confised on this question overall. i got up to finding moles and timesing by 10, but didnt know how to use these numbers to find the mass of anhydrous sodium carbonate?

the question:
A student added 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) to 200
cm3 of 0.250 mol dm–3 hydrochloric acid in a beaker and stirred the mixture.
After the reaction was complete, the resulting solution was transferred to a
volumetric flask, made up to 250 cm3 with deionised water and mixed thoroughly.
Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol
dm–3 aqueous sodium hydroxide. The mean titre was 26.60 cm3 of aqueous
sodium hydroxide.
Calculate the value of x in Na2CO3.xH2O
Show your working.
Give your answer as an integer.

the mark scheme:
M1 HCl added = 0.050 mol and
NaOH used in titration = 3.99 × 10–3 mol
1
M2 So moles that would be needed to neutralise total excess
HCl = 3.99 × 10–3 × 10 = 3.99 × 10–2 mol

1
M3 Therefore the moles of HCl reacted with the Na2CO3.xH2O =
0.050 - 3.99 × 10–2 = 0.0101 mol

M4 So moles Na2CO3.xH2O reacted with the HCl = 0.0101 / 2 = 5.05 x
10–3 mol
1
M5 Conversion of mg to g = 0.627 (g) or 627 × 10–3
(g)
1
M6 xH2O = 0.627/5.05 × 10–3
-106.0 = 18 (.16)

1
M7 so x = 1
Alternative: 0.0917 /18.0 = 5.094 × 10–3 so ratio

To solve this problem, we need to use the balanced chemical equation for the reaction of sodium carbonate with hydrochloric acid and the data from the titration to determine the number of moles of sodium carbonate that were present in the original mixture.

The balanced chemical equation for the reaction of sodium carbonate with hydrochloric acid is:

Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2

Since the reaction consumes 1 mole of hydrochloric acid for every mole of sodium carbonate that reacts, we can use the volume and concentration of the hydrochloric acid and the volume and concentration of the aqueous sodium hydroxide to determine the number of moles of sodium carbonate that were present in the original mixture.

First, we can calculate the number of moles of hydrochloric acid that were present in the original mixture:

number of moles of HCl = (200cm3)(0.250mol/dm3) = 0.0500 mol

Then, we can calculate the number of moles of sodium hydroxide that were consumed in the titration:

number of moles of NaOH = (26.60cm3)(0.150mol/dm3) = 0.0395 mol

Since the reaction consumes 1 mole of hydrochloric acid for every mole of sodium carbon.

Let me know if you'd like more help on any other questions that you have got!

thank you!!

so first you find the moles of HCl in the original solution - A , and the moles of NaOH consumed in the titration - B

then because the titration was done for a 25cmcubed sample of the original volume of the solution, you do Bx10 = C beause 25x10 is 250

now because there are 2 reactions going on - the titration and HCl with Na2CO3 - you find out how many moles of HCl were used up in the titration, and so how many moles are left to react with the Na2CO3. so you do A-C = D mol

then you can find the moles of Na2CO3 in the original solution by using the mole ratios in the balanced eqn, so you do D/2 = E mol of Na2CO3

then use moles x mr of ONLY Na2CO3 (not including h2o) = mass

then do the mass of all the hydrated na2co3 minus the mass of only na2co3 to find the mass of h2o only

and find the ratios to find x

i think i was confused bc i didnt know what was reacting with what.