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    what volume of 0.05mol dm-3 KOH is required to exactly neutralise 24.0 cm3 of 0.11 mol dm-3 H2SO4 according to the equation 2KOH +H2SO4= K2SO4+2H2O??
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    First work out the number of moles of H2SO4 that you have.

    0.11 moldm-3 * 24.0 * 10^-3 dm3 = 2.64*10^-3 moles of H2SO4.

    Each mole of H2SO4 reacts with two moles of KOH (just look at the equation)

    Therefore 5.28*10^-3 moles will react.

    5.28*10^-3 moles / 0.05 moles dm-3 = 105.6 * 10^-3 dm3

    This is 105.6 cm3.

    The key to a lot of these questions is just to use standard form of some description. I use engineering notation which involves powers of three only, as this makes unit conversions straightforward. Just remember to always replace dm3 with * 10^-3 cm3 and you'll be fine.
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    I just attempted this question and got an answer a bit different. i worked out the number of moles of h2so4 and converted this answer to dm3 by dividing this by the concentration(since volume=moles/concentration). since this volume is only for one mole of h2so4 I multiplied this by 2(to get 2 moles of koh).

    my answer was 0.1078 dm3, is this right?


    Posted from TSR Mobile
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    (Original post by zaaraxo)
    I just attempted this question and got an answer a bit different. i worked out the number of moles of h2so4 and converted this answer to dm3 by dividing this by the concentration(since volume=moles/concentration). since this volume is only for one mole of h2so4 I multiplied this by 2(to get 2 moles of koh).

    my answer was 0.1078 dm3, is this right?


    Posted from TSR Mobile
    We are within 2% so the difference will just be down to rounding. That's a new method I had never thought of, will have to try it.

    Posted from TSR Mobile
 
 
 
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