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# Hydrated Salt Question watch

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1. Hi, I'm stuck on this question..I can't seem to get an appropriate answer for x, can you guys have a go?

Compound is X2CO3.10H2O
Mass of hydrated carbonate = 2.01g
Mass of anhydrous carbonate = 0.74g
Mass of H20 driven off = 1.27g

Q1). Calculate the amount, in moles, of H2O removed from the sample of the hydrated salt when heated.

Q2). Construct an equation for the reaction

Q3). Deduce the moles of the anhydrous salt

Q4). calculate the Mr of the anhydrous salt and hence the value of X

I keep getting 3.1gmol-1 or 22.441gmol-1..... what do you guys get?
2. Where are you stuck?
3. The answers i'm getting don't seem right

I think the problem stems from Q1
4. Q1). Calculate the amount, in moles, of H2O removed from the sample of the hydrated salt when heated.

Moles=Mass/Mr
Moles=1.27/18
Moles=0.0705

Q3). Deduce the moles of the anhydrous salt

Moles ratio of H2O to the salt:
10:2
Therefore, divide by 5:
0.0705/5=0.0141
Moles of salt= 0.0141

I've had a go at question 1 and 3. Let me know if it's correct or not.
5. I'll show you what I've done

Q1 n=m/Mr 1.27/18 = 0.07056 Mol

X2CO3.10H20 -----> X2CO3 + 10H2O

which shows a 1:10 ratio
therefore divide by 10

0.07056/10 = 0.007056 Mol (of X2CO3)

Mr=m/n 0.74/0.007056 = 104.875

104.875 - 60 (to work out X2) = 44.875

44.875/2 = 22.4
x=22.4

but what metal has an Ar of 22.4.....hence why i think i've gone wrong
6. (Original post by grahammnmr)
which shows a 1:10 ratio
therefore divide by 10
Wouldn't the mole ratio be 2:10 since its 2CO3?
7. (Original post by derpz)
Wouldn't the mole ratio be 2:10 since its 2CO3?
X2CO3

2 as in two lots of 'x', sorry my fault with the font sizes
8. I get 22.48 as well. Sodium does have an isotope of mass 22.

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