During the decomposition of `H_(2)O_(2)` to give oxygen, `48 g O_(2)` is formed per minute at a certain point of time. The rate of formation of water at this point is

Options

A. `0.75 "mol min"^(-1)`
B. `1.5 "mol min"^(-1)`
C. `2.25 "mol min"^(-1)`
D. `3.0 "mol min"^(-1)`

Answer

D

Explanation

`2H_(2)O_(2)rarr2H_(2)O+O_(2)`
Rate of formation of `O_(2)=((48g)/(32"g mol"^(-1)))"min"^(-1)`
`=(3)/(2)"mol min"^(-1)=1.5 "mol min"^(-1)`
Rate of formation of water
`=2xx` Rat of formation of `O_(2)`
`=2xx1.5 "mol min"^(-1)`
`=3 "mol min"^(-1)`.