A Sample Of KCl3 On Decomposition Yielded 448 Ml Of Oxygen Gas At Ntp.Calculate (i) Weight Of Oxygen Product (ii) Weight Of KClO3 Originally Taken (iii) Weight Of KCl Produced.

2KClO3 \(  \overset{\Delta }{\rightarrow}  \)2KCl + 3O2.

Number of moles of Oxygen (O2)  at STP = \( \frac{V(ml)}{22400} \).

\( \Rightarrow \frac{448}{22400} \).

= 0.02 mole.

Mass of 1 mole of Oxygen (O2) = 32 gms .

Mass of 0.02 mole of Oxygen (O2) = 32 * 0.02 = 0.64 gms .

Mass of KClO3 = moles(n)  *  molecular weight .

= 2  *  122.5 = 245 gms .

Weight of KClO3 to produce 3 moles of Oxygen (O2)  = \( \frac{ 245  * 0.02 }{3} \).

=  1.63 gms

Mass of KCl produced = Mass of KClO – Mass of O2 produced

= 1.63 − 0.64 = 0.993 gms.

(i) Weight of Oxygen (O2)  produced = 0.64 gms .

(ii) Weight of  KClO3 originally  taken = 1.63 gms.

(iii) Weight of KCl produced = 0.993 gms .

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