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Size:  503.0 KBCan someone tell me how to get b(iii)?
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    If the handwriting on the paper is your, you have made a good start by calculating the number of moles of NaOH that reacted.

    SO2 and HCl are made in the initial reaction in equal number of moles. This is key.

    Now look at the 2 equations for reactions with NaOH.
    For every 3 moles of NaOH that are neutralised in the titration:
    1 mole NaOH reacts with 1 mole of HCl
    2 moles NaOH react with 1 mole of SO2

    Therefore for every 3 moles of NaOH titrated, 1 mole HCl + 1 mole SO2 are neutralised, i.e. 2 moles in total.
    So the number of moles of HCl + SO2 neutralised = 2/3 the number of moles of NaOH titrated.
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    (Original post by TutorsChemistry)
    If the handwriting on the paper is your, you have made a good start by calculating the number of moles of NaOH that reacted.

    SO2 and HCl are made in the initial reaction in equal number of moles. This is key.

    Now look at the 2 equations for reactions with NaOH.
    For every 3 moles of NaOH that are neutralised in the titration:
    1 mole NaOH reacts with 1 mole of HCl
    2 moles NaOH react with 1 mole of SO2

    Therefore for every 3 moles of NaOH titrated, 1 mole HCl + 1 mole SO2 are neutralised, i.e. 2 moles in total.
    So the number of moles of HCl + SO2 neutralised = 2/3 the number of moles of NaOH titrated.
    I understand how you got the 2/3 mol of HCl + SO2 but why is the mol divided by 3?
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    (Original post by Snowie9)
    I understand how you got the 2/3 mol of HCl + SO2 but why is the mol divided by 3?
    (Original post by Snowie9)
    I understand how you got the 2/3 mol of HCl + SO2 but why is the mol divided by 3?
    Because for every 2 moles HCl +SO2 created, 3 moles NaOH will be titrated.
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    (Original post by TutorsChemistry)
    Because for every 2 moles HCl +SO2 created, 3 moles NaOH will be titrated.
    I'm still a little bit confused in why mol of NaOH divided by 3 will be equal to mol of acid ><
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    It's not 3 we need to divide by. We multiply by 2/3.

    For every 2 moles of acidic gas generated in our reaction, we require 3 moles of NaOH to netralise them.

    You have calculated the number of moles of NaOH used in the neutralisation. Multiply this number of moles by 2/3 to get the total number of moles of the acidic gases generated.
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    (Original post by TutorsChemistry)
    It's not 3 we need to divide by. We multiply by 2/3.

    For every 2 moles of acidic gas generated in our reaction, we require 3 moles of NaOH to netralise them.

    You have calculated the number of moles of NaOH used in the neutralisation. Multiply this number of moles by 2/3 to get the total number of moles of the acidic gases generated.
    Oh i got the answer after i multiply it by 2/3 and then divide it by 2. Thanks
 
 
 
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