Correct Answer - C
Calculation of `O_(2)` gas required to burn 112 L of CO.
`"CO"+1/2 "O"_(2) rarr"CO"_(2)`
`1 "mol ""0.5 mol ""1 mol" `
=22.4 L
`22.4` L of CO at NTP require `O_(2) = 0.5` mol
112 L of CO at NTP will require `O_(2)=(0.5)/(22.4)xx 112 = 2.5` mol
This `O_(2)` is to be obtained by heating `KClO_(3)`.
`underset(2" mol")(2KC)lO_(3) rarr 2KCl +underset(3" mol")(3O_(2))`
`2(39+35.5 3 xx 16) = 245` g
3 moles of `O_(2)` are produced from `KClO_(3)=245`g
2.5 moles of `O_(2)` will be produced from
`KClO_(3)=(245xx2.5)/3=204.167` g