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A-level Chemistry HELP !! watch

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    (b) A 0.11 g sample of pure barium was added to 100 cm3 of water.
    Ba(s) + 2H2O(l) Ba(OH)2(aq) + H2(g)

    (i) Show that 8.0 × 10−4 mol of Ba were added to the water.
    [1]
    0.11/137.3= 8x10-4



    (ii) Calculate the volume of hydrogen, in cm3, produced at room temperature and pressure.


    (iii) Calculate the concentration, in mol dm−3, of the Ba(OH)2(aq) solution formed.


    HOW DO YOU DO ii and iii ?
    thanks
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    someone ?
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    At Standard Temperature Pressure (S.T.P.) 1 Mole of any gas occupies 22.4 cm3. This is equal to 22.4 Litres.

    P.S. I'm only a GCSE student
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    Looking at the balanced equation, you know that one mole of barium will produce one mole of hydrogen gas when it is reacted with water.

    Using this same idea with the barium hydroxide, you have a volume and a number of moles for barium hydroxide.

    C = N/V
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    (Original post by chem.man)
    (b) A 0.11 g sample of pure barium was added to 100 cm3 of water.
    Ba(s) + 2H2O(l) Ba(OH)2(aq) + H2(g)

    (i) Show that 8.0 × 10−4 mol of Ba were added to the water.
    [1]
    0.11/137.3= 8x10-4



    (ii) Calculate the volume of hydrogen, in cm3, produced at room temperature and pressure.


    (iii) Calculate the concentration, in mol dm−3, of the Ba(OH)2(aq) solution formed.


    HOW DO YOU DO ii and iii ?
    thanks
    For part (iii) you need to use n=C x V and the stoichiometric ratio given in the equation (which you already have)
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    ii) 8.0 x 10-4 x 24 dm3 = 0.0192 dm3
    0.0192 x 1000 = 19.2 cm3

    iii) moles Ba = moles Ba(OH)2
    8 x 10-4 / 100/1000 = 8x10-3 mol dm-3
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    Thanks all !
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    (Original post by pineneedles)
    Looking at the balanced equation, you know that one mole of barium will produce one mole of hydrogen gas when it is reacted with water.

    Using this same idea with the barium hydroxide, you have a volume and a number of moles for barium hydroxide.

    C = N/V
    how did you know that you have to use the rato between Ba and H2 and not the ratio of 2H2O and H2 ?

    thanks.
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    Use the stoichiometric ratio for ii and then use the formula moles = volume/molar volume and use c = m/v for iii
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    (Original post by chem.man)
    how did you know that you have to use the rato between Ba and H2 and not the ratio of 2H2O and H2 ?

    thanks.
    You need to use the ratio between Ba and H2 because you're looking for the volume of hydrogen and you know the amount of Ba that reacted. The amount of Ba also limits the reaction (likely so) which is another reason why you need to use that value. Have you come across the idea of a limiting reactant before?
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    (Original post by pineneedles)
    You need to use the ratio between Ba and H2 because you're looking for the volume of hydrogen and you know the amount of Ba that reacted. The amount of Ba also limits the reaction (likely so) which is another reason why you need to use that value. Have you come across the idea of a limiting reactant before?
    ah ok , and yes , but I have only covered it in GCSE not in AS yet
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    (Original post by chem.man)
    ah ok , and yes , but I have only covered it in GCSE not in AS yet
    If you can appreciate that a reaction is limited by the reactant which you have fewer moles of, then you're fine. I'm sure that idea doesn't change at all in AS
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    (Original post by pineneedles)
    If you can appreciate that a reaction is limited by the reactant which you have fewer moles of, then you're fine. I'm sure that idea doesn't change at all in AS

    thank you !! , honestly , your help was really valued

    appreciate it
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    (Original post by AngryRedhead)
    Right; you need to use pV=nRT for part (ii) and for part (iii) you need to use n=C x V and the stoichiometric ratio given in the equation (which you already have)
    okay , but for part ii) what will the number of moles be for H2 ?
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    (Original post by chem.man)
    okay , but for part ii) what will the number of moles be for H2 ?


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    Look at the moles ratio in your balanced equation.
 
 
 
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