Your reaction will produce #"0.94 L"# of oxygen gas.
The first thing you need to do is determine how many moles of oxygen gas are produced when that much silver is produced. To do this, use the number of moles of silver and the mole ratio that exists between the two products.
#2Ag_2O_((s)) -> color(red)(4)Ag_((s)) + O_(2(g))#
Notice that you have a #color(red)(4):1# mole ratio between silver and oxygen, which means that you get 1 mole of oxygen for every 4 moles of silver produced.
The number of moles of silver the reaction produced is
#15.9cancel("g") * "1 mole Ag"/(107.87cancel("g")) = "0.1474 moles Ag"#
This means that you'll get
#0.1474cancel("moles Ag") * ("1 mole "O_2)/(4cancel("moles Ag")) = "0.03685 moles "# #O_2#
Now for the tricky part. Your oxygen gas was collected over water at a total pressure of 757 mmHg. This pressure includes the pressure of the water vapor at that temperature (#25^@"C"#).
The vapor pressure of water at #25^@"C"# is 26.69 mmHg (http://www.endmemo.com/chem/vaporpressurewater.php), which means that the pressure of the oxygen will be
#P_"total" = P_"water" + P_(O_2) => P_(O_2) = P_"total" - P_"water"#
#P_(O_2) = 757 - 26.69 = "730.3 mmHg"#
Now you've got all the values you need to apply the ideal gas law equation
#PV = nRT => V = (nRT)/P#
#V = (0.03685cancel("moles") * 0.082(cancel("atm") * "L")/(cancel("mol") * cancel("K")) * (273.15 + 25)cancel("K"))/(730.3/760cancel("atm"))#
#V = "0.9375 L"#
Rounded to two sig figs, the answer will be
#V_(O_2) = color(green)("0.94 L")#
