The answer is #32.4^@C#.
First, start with the balanced chemical equation
#2Na_2O_2(s) + 2H_2O_((l)) -> 4NaOH_((s)) + O_2(g)#
In order to determine the heat of reaction, #DeltaH#, we need the standard formation enthalpy values (#DeltaH_f^@#) for the reactants and the products and the number of moles that react
#Na_2O_2#: #DeltaH_f^@ = -504.6 (kJ)/(mol)#;
#H_2O#: #DeltaH_f^@ = -285.8 (kJ)/(mol)#;
#NaOH#: #DeltaH_f^@ = -470.1 (kJ)/(mol)#;
#O_2#: #DeltaH_f^@ = 0 (kJ)/(mol)#.
We know that #Na_2O_2#'s molar mass is #78.0 g/(mol)#, which means that the number of moles of #NaOH# is
#n_(Na_2O_2) = m_(Na_2O_2)/(molarmass) = (7.800 g)/(78.0 g/(mol)) = 0.100# moles
We can determine the number of molesof water by using its density of approximately #1.00 g/(mL)#, its molar mass of #18.0 g/(mol)#, and its given volume
#n_(H_2O) = m_(H_2O)/(molar mass) = (rho * V)/(molar mass) -> #
#n_(H_2O) = (1.00 g/(mL) * 110.00 mL)/(18.0 g/(mol)) = 6.11# moles
We can see that #Na_2O_2# is the limiting reagent, which means that the number of moles of water that will react will be #n_(H_2O) = n_(Na_2O_2) = 0.100#.
Therefore, #DeltaH# can be calculated to be (keeping in mind the mole-to-mole ratios)
#DeltaH = 0.200 * (-470.1) - (0.100 * (-285.8) + 0.100 * (-504.6)) = --14.9kJ#
Since #DeltaH<0#, we are dealing with an exothermic reaction; this means that the heat absorbed by the water will be #q = -DeltaH = 14.9kJ#. The change in temperature can then be calculated to be ( #c_(H_2O)# - water's specific heat):
#q = m_(H_2O) * c_(H_2O) * DeltaT -> DeltaT = q/(m * c)#
#DeltaT = (14.9 * 10^3 J)/(110.00g * 4.18 J/(g^@C)) = 32.4^@C#
The temperature of the water increased by #32.4^@C#.
