For the reaction `4NH_(3) + 5O_(2) to 4NO + 6H_(2) O`, if the rate of disappearance of `NH_(3)` is `3.6 xx 10^(-3)` mol `L^(-1) s^(-1)`, what is the rate of formation of `H_(2) O`

Options

A. `5.4 xx 10^(-3) mol L^(-1) s^(-1)`
B. `3.6 xx 10^(-3) mol L^(-1) s^(-1)`
C. `4 xx 10^(-4) mol L^(-1) s^(-1)`
D. `0.6 xx 10^(-4) mol L^(-1) s^(-1)`

Answer

A

Explanation

`- (1)/(4) (d[NH_(3)])/(dt) = + (1)/(6) (d[H_(2)O])/(dt)`
`(d[H_(2)O])/(dt) = (6)/(4) xx 3.6 xx 10^(-3) = 5.4 xx 10^(-3) mol L^(-1) s^(-1)`