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Need help with chemistry volumetric calculation questions watch

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    Name:  1447356054914-385745708.jpg
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Size:  383.9 KB can anyone help me with the questions circled in red? I don't know what to work out the vol of the resulting product and have been getting the wrong answers. Thanks
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    Attachment 477555477557 for q3 and 4 what does it mean by excess acid?? The volume of the excessive acid isn't specified.. Thanks
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    (Original post by coconut64)
    Attachment 477555477557 for q3 and 4 what does it mean by excess acid?? The volume of the excessive acid isn't specified.. Thanks
    You use the NaOH to neutralise the acid left over from the first reaction.

    moles of NaOH in the titration = moles of excess acid.
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    (Original post by charco)
    You use the NaOH to neutralise the acid left over from the first reaction.

    moles of NaOH in the titration = moles of excess acid.
    For 3, the number of mole of naoh is 0.03 and I did 0.03/01 then ×1000 but this doesn't give me the right answer which is 25cm3
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    (Original post by coconut64)
    For 3, the number of mole of naoh is 0.03 and I did 0.03/01 then ×1000 but this doesn't give me the right answer which is 25cm3
    In #3 mol NaOH = 20/40 = 0.5 mol

    this is in 200ml

    hence concentration = 0.5/0.2 = 2.5 mol dm-3
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    (Original post by coconut64)
    For 3, the number of mole of naoh is 0.03 and I did 0.03/01 then ×1000 but this doesn't give me the right answer which is 25cm3
    ahhh you mean no 3 on the second sheet.

    25ml of 0.5 mol dm-3 sulfuric acid = 0.025 * 0.5 mol = 0.0125 mol

    This is mixed with 0.03 mol NaOH

    equation for the reaction is:

    2NaOH + H2SO4 --> Na2SO4 + 2H2O

    i.e 0.0125 mol H2SO[sub]4[/sub reacts with 0.025 mol NaOH

    This means that there are 0.03 - 0.025 = 0.005 mol NaOH excess

    can you take it from there?
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    (Original post by charco)
    ahhh you mean no 3 on the second sheet.

    25ml of 0.5 mol dm-3 sulfuric acid = 0.025 * 0.5 mol = 0.0125 mol

    This is mixed with 0.03 mol NaOH

    equation for the reaction is:

    2NaOH + H2SO4 --> Na2SO4 + 2H2O

    i.e 0.0125 mol H2SO[sub]4[/sub reacts with 0.025 mol NaOH

    This means that there are 0.03 - 0.025 = 0.005 mol NaOH excess

    can you take it from there?
    How did you get o.025?
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    (Original post by charco)
    In #3 mol NaOH = 20/40 = 0.5 mol

    this is in 200ml

    hence concentration = 0.5/0.2 = 2.5 mol dm-3
    why did you do 20/40 here? What happens to the mole of naoh as water s added. By ml do you mean mol here? Thank you
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    (Original post by coconut64)
    why did you do 20/40 here? What happens to the mole of naoh as water s added. By ml do you mean mol here? Thank you
    This is no. 3 in sheet 1
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    (Original post by coconut64)
    How did you get o.025?
    From the balanced equation. If you have 0.0125 mol of sulfuric acid it will react with 0.025 mol of NaOH
 
 
 
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