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    "50cm3 of 0.1mol dm-3 solution of sulphuric acid is to be neutralised by KOH as indicated by the equation below

    H2SO4 + 2KOH → K2SO4 + 2H2O

    what volume of 0.05 moldm-3 KOH will be required?"
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    (Original post by tomlfc)
    "50cm3 of 0.1mol dm-3 solution of sulphuric acid is to be neutralised by KOH as indicated by the equation below
    H2SO4 + 2KOH → K2SO4 + 2H2O
    what volume of 0.05 moldm-3 KOH will be required?"
    This is not a hard question.

    You know the reaction takes place in a 1:2 ratio.

    Convert your sulphuric acid into mols, adjust to the ratio above, and convert back into the desired volume.
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    (Original post by Fango_Jett)
    This is not a hard question.

    You know the reaction takes place in a 1:2 ratio.

    Convert your sulphuric acid into mols, adjust to the ratio above, and convert back into the desired volume.
    how would i convert sulphuric acid into mols though?
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    (Original post by tomlfc)
    how would i convert sulphuric acid into mols though?
    You know the volume and concentration... n = v x c
 
 
 
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