[KClO_(4)" can be prepared by following reaction sequence : "],[" (i) "Cl_(2)+2KOH rarr KCl+KClO+H_(2)Oquad " (ii) "3KClO rarr2KCl+K]

KClO_(4) can be prepared by following reactions: i. Cl_(2) + 2 KOH rarr KCl + KClO + H_(2) O ii. 3KClO rarr 2KCl + KClO_(3) iii. 4KClO_(3) rarr 3KlI_(4) + KCl (Atomic weight of K, Cl , and O are 369,35.5 and 16)

KClO_(4) can be prepared by following reactions: i. Cl_(2) + 2 KOH rarr KCl + KClO + H_(2) O ii. 3KClO rarr 2KCl + KClO_(3) iii. 4KClO_(3) rarr 3KlI_(4) + KCl (Atomic weight of K, Cl , and O are 369,35.5 and 16)

KClO_(4) can be prepared by the following reaction: Cl_(2)+2KOH to KCl + KClO+H_(2)O 3KClO to 2KCl +KClO_(3), 4KClO_(3) to 3KClO_(4)+KCl To prepare 200g KClO_(4) , the required amount of Cl_(2) is equivalent to-

Complete the following reactions : KClO_(3) rarr ?

Balance the following chemical equations. d) KClO_(3)rarr KCl +O_(2)

Which catalyst are used in following reactions : 2KClO_3 to 2KCl + 3O_2

In the following disproportionation of Cl_(2) in basic medium Cl_(2)+2KOH rarr KCl+KClO+H_(2)O Equivalent mass of Cl_(2) is

Balance the following equation : KOH + Cl_2 to KCl + KClO + H_2O .