Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation. 6 OH - + Cl - → ClO 3 - + 3 H 2 O + 6 e - A current of xA has to be passed for 10 h to produce 10 . 0 g of potassium chlorate. the value of x is __________. (Nearest integer) (Molar mass of KClO 3 = 122 . 6 g mol - 1 F = 96500 C )

Step-by-step solution

Given balanced equation is 60 H ⊝ + C ⊝ l ⟶ ClO 3 - + 3 H 2 O + 6 e - → 10   gKClO 3 ⇒ 10 122 . 6   molKCO 3 in obtained → from the above reaction, it is concluded that by 6   F charge 1   molKClO 3 is obtained. → By the passage of 6   F charge = 1   molKClO 3 ∴ By the passage of x × 10 × 60 × 60 96500 F charge = 1 6 × x × 10 × 60 × 60 96500 Now x × 10 × 60 × 60 6 × 96500 = 10 122 .