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# Concentration of standard iodine solution...? watch

1. delete
2. 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
3. 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
4. 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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