Question

How many grams of iodine (I2) are produced when 32.00 ml of 0.115 M NaOCl reacts with 25.0 ml of 0.175 M KI (and excess H+ )?

Answer 1

→ The objective is to how many gram, determine produced. of lodine (92) are No.of moles of Naoc 0 .115m 32.00 ML Naocel - volume of No.of moles volume of of KI k? = 0.145M 25.0m 1 . Naoci 32 * 0.115 millmoles of 3.68 millimoles of 25 * 0.145 4.375 Balanced full equation : Nacl (aq) + 2 KOH (aq) +12/) Naocllaq) + PKI (29H H20-→ of limiting agent Nacel consumed 4.395 KI = of 2 mdee by 4.375 moles of 62. ki will give 1 millind of I2. 2 millimoles of So, 4•375 of hy will give 4.375 millmoles millimoles of 12. ili millimoles of 2.1875 72.

of I, millimolee 2.1875 mdar mass of I, = 253.8 So, mass of I, - (millimales of 12 ) ( moler mass $1.) = 62.18454103) ( 253.8 ) .oosssgm. I2 = 0.555 gm I mass of *