The answer is 32.4^@C32.4∘C.
First, start with the balanced chemical equation
2Na_2O_2(s) + 2H_2O_((l)) -> 4NaOH_((s)) + O_2(g)2Na2O2(s)+2H2O(l)→4NaOH(s)+O2(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.
