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    So I've been stuck on this AS Level CH2 Past paper question for too long and I would be very, very grateful if anyone could please help me with it. Thanks! Here it is:

    Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H2O, is formed.
    CaCl2 (s) + 2H2O --> CaCl2.2H2O
    Mr: 111
    In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, was used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)
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    (Original post by Gwenog_quidditch)
    So I've been stuck on this AS Level CH2 Past paper question for too long and I would be very, very grateful if anyone could please help me with it. Thanks! Here it is:

    Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H2O, is formed.
    CaCl2 (s) + 2H2O --> CaCl2.2H2O
    Mr: 111
    In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, was used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)
    Moles of cacl2 = 5.55/111 = 0.05 mol
    Moles of H2O that can be removed= 0.1 mol
    Mass of H20 THAT can be removed = 0.1 x 18 = 1.8g
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    (Original post by thebrahmabull)
    Moles of cacl2 = 5.55/111 = 0.05 mol
    Moles of H2O that can be removed= 0.1 mol
    Mass of H20 THAT can be removed = 0.1 x 18 = 1.8g
    Thanks,where did you get the 0.1 mol for H2O from though?
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    (Original post by thebrahmabull)
    Moles of cacl2 = 5.55/111 = 0.05 mol
    Moles of H2O that can be removed= 0.1 mol
    Mass of H20 THAT can be removed = 0.1 x 18 = 1.8g
    Never mind, I know now thanks so much
 
 
 
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