text: a sample of 0.6013 g of potassium iodate was dissolved in water. hydrochloric acid and solid potassium iodide? were then added in excess. what is the minimum mass of of solid ki and the minimum volume of 3.00 m hcl required to convert all the ions io3– to i3– ions?  kio3– (aq)  +  8ki–  + 6hcl(aq)   -------->  3ki3– (aq) + 3h20 + 6kcl 0.6013 g kio3*(1mol kio3/ 214g kio3)= 0,002810 moles of kio3 (aq)  one mol of kio3 (aq) reacts with 8 moles of ki and 6 moles of hcl. therefore 0,002810 moles kio3*(8 moles ki/1mol kio3)(166g ki/1mol ki)= 3,732 g of ki  for hcl 0,002810 moles kio3*(6 moles hcl/1mol kio3)(36,45 g hcl /1mol ki) = 0,6145 grams of hcl = 0,01686 moles hcl because you have your hcl dissolved, then  3,00 m = moles of hcl/volume of solution 3,00 m = 0,01686 moles /? ? = 5,62 ml of hcl (aq)