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    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?
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    Where are you stuck?
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    The answers i'm getting don't seem right

    I think the problem stems from Q1
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    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.
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    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
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    (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?
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    (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
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    I get 22.48 as well. Sodium does have an isotope of mass 22.
 
 
 
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