During the decomposition of H₂O₂ to give oxygen, 48 g O₂ is formed per minute at a certain point of time. The rate of formation of water at this point is:

3.0 mol min^–1

Here, the given reaction is

$2H_2O_2 −−−−→ 2H_2O  +  O_2$

Rate $= − \frac{1}{2}\frac{d[H_2O_2]}{dt} = \frac{1}{2}\frac{d[H_2O]}{dt} = \frac{d[O_2]}{dt}$

Rate of formation of oxygen $= 48 \text{g min}^{−1}$

                                          $= \frac{48}{32}\text{ mol min}^{−1}$

                                           $= 1.5 \text{ mol min}^{−1}$

Rate of formation of $H_2O$

$= \frac{1}{2}\frac{d[H_2O]}{dt} = \frac{d[O_2]}{dt}$

$= \frac{d[H_2O]}{dt} = \frac{2d[O_2]}{dt}$

$= \frac{d[H_2O]}{dt} = 2 × 1.5 \text{ mol min}^{−1}$

$= \frac{d[H_2O]}{dt} = 3 \text{ mol min}^{−1}$