$1{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O$${R_{{H_2}O}} = 2 \times 2/4\frac{{mol}}{{L.\min }} = \frac{{\frac{{14/4}}{{18}}}}{{2 \times \Delta t}}$$\Delta t(s) = \frac{{\frac{{14/4}}{{18}}}}{{2 \times 2 \times 2/4}} \times 60 = 5s$