Answer
$0.0963 mols H_{2}O_{(l)}$
Work Step by Step
Using the ideal gas law $PV = nRT$ where "P" is the pressure the gas exerts, "V" is the volume the gas occupies, "n" represents the number of moles of the gas, "R" is the ideal gas constant and "T" represents the temperature, we can solve many gaseous questions that behave in ideal conditions.
The question asks for $n$ so we must isolate for the volume from the ideal gas equation to obtain $n = \frac{PV}{RT}$
$n= \frac{0.988 atm \times 1.3 L }{325 K\times 0.08206 \frac{L\times atm}{mol\times K}} = 0.0482 mols O2 $
$ 2H_{2}O_{(l)} --> 1O_{2}._{(g)} + 2H_{2}._{(g)}$
Since the balanced equation shows a 2:1 mole ratio between $H_{2}O_{(l)}$ and $O_{2}._{(g)}$, twice the number of mols of $H_{2}O_{(l)}$ must react to form one mol of $O_{2}._{(g)}$ so in this case; $2*0.0482 =0.0963 mols H_{2}O_{(l)}$
