12 A student adds 3 mol of acidified K 2Cr 2O7 to an excess of I– ions.The chromium is all reduced to Cr 3+ and I– ions are oxidised to I2.The I2 released is reduced back to I– ions by X mol of S 2O32– ions.1 mol of I2 is reduced by 2 mol of S 2O32– ions.What is the value of X?A 3 B 6 C 9 D 18

The reaction between K2Cr2O7 and I- ions results in the formation of Cr3+ and I2. According to the question, all the I2 formed is then reduced back to I- ions by S2O32- ions.

The stoichiometry of the reaction between I2 and S2O32- ions is given as 1 mol of I2 is reduced by 2 mol of S2O32- ions.

Since 3 mol of K2Cr2O7 are used, and the reaction between K2Cr2O7 and I- ions produces 1 mol of I2 for each mol of K2Cr2O7, this means that 3 mol of I2 are produced.

Therefore, to reduce these 3 mol of I2 back to I- ions, 2 * 3 = 6 mol of S2O32- ions are needed.

So, the value of X is 6. The correct answer is B. 6.