Question

Oxygen is prepared by catalytic decomposition of potassium chlorine `(KClO_(3))`. Decomposition of potassium, chloride gives potassium chloride (KCl) and oxygen `(O_(2))`. How many moles and how many grams of `KClO_(3)` are required to produce 2.4 mole `O_(2)`?

Answer 1

Decomposition of `KClO_(3)` takes places as,
`2KClO_(3)(s) to 2KCl(s)+3O_(2)(g)`
`2"mole" of KClO_(3)-=3"mole of"O_(2)`
`therefore "3 mole "O_(2)"formed by 2 mole" KClO_(3)`
`therefore 2,4 "mole" O_(2) "will be formed by" ((2)/(3)xx2.4)"mole"KClO_(3)`
`=1.6"mole of" KClO_(3)`
Mass of `KClO_(3)`= Number of moles `xx` Molar mass
`=1.6xx122.5=196g`