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    Work out how many moles of KIO3 and KI you will be adding, then which is ever the smaller amount is the limiting reagent. Multiply this value by 3 to work out the number of I2 moles which are formed, then divide by the volume
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    Cheers, ive always had difficulty with the coefficient. So finding the moles of KI would look like this. I can solve the rest, think I just had a memory blank >.<

    1.6/165.998 = 0.009639 moles of 5KI

    0.009639/5 = 0.001928 moles of KI


    PS. KIO3 was the LR so I worked it all out now... thanks for help man.. wow im an idiot sometimes
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    So there are 0.009639 moles of 5KI

    and 0.009439 moles of KIO3

    so the maximum amount of I2 produced is 0.0283 mol
 
 
 
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