text: 0.6013 g KIO3*(1mol KIO3/ 214g KIO3)= 0,002810 moles of KIO3 (aq)\n\nOne mol of KIO3 (aq) reacts with 8 moles of KI and 6 moles of HCl.\nTherefore\n0,002810 moles KIO3*(8 moles KI/1mol KIO3)(166g KI/1mol KI)=\n3,732 g of KI\n\nFor HCl\n0,002810 moles KIO3*(6 moles HCl/1mol KIO3)(36,45 g HCl /1mol KI) = 0,6145 grams of HCl = 0,01686 moles HCl\nBecause you have your HCl dissolved, then\n\n3,00 M = moles of HCl/volume of solution\n3,00 M = 0,01686 moles /?\n? = 5,62 mL of HCl (aq)