Electrolysis of Water :)

The balanced reaction of decomposition of water is given by:

$\ce{2H2O (l) -> 2H2 (g) + O2 (g)}$

At same temperature and pressure, volume of a gas is proportional to its moles. Therefore, the volume of $\ce{H2}$ produced in this case is twice that of $\ce{O2}$ or $13.44 \times 2 = \text{26.88 L}$ .

At STP, $\text{1 mol}$ of gas has a volume of $\text{22,4 L}$ , therefore, $\text{13.44 L}$ of $\ce{O2}$ is $\dfrac{13.44}{22.4} = \text{0.6 mol}$ . Since $\text{2 mol}$ of $\ce{H2O}$ produces $\text{1 mol}$ of $\ce{O2}$ . Therefore, the mole of $\ce{H2O}$ decomposed was $0.6 \times 2 = \text{1.2 mol}$ $= 1.2 \times (1\times 2 + 16)$ $=1.2 \times 18 = \text{21.6 g}$ .

The answer is $\boxed{\text{21.6 g, 26.88 L}}$