Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.
The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,
total amount of heat = 13.0 g (2.09 J/gC)(12) + 13(333.55 J/g) + 13.0 g (4.18 J/gC)(100 - 0) + (13.0 g)(2260 J/g) + (13 g)(2.01 J/g)(113-100)
= 39815.88 J
= 39.82 kJ
