Answer: The enthalpy of the reaction for given amount of [tex]Na_2O_2[/tex] will be -8.064 kJ.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of [tex]Na_2O_2[/tex] = 10 g
Molar mass of [tex]Na_2O_2[/tex] = 78 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }Na_2O_2=\frac{10g}{78g/mol}=0.128mol[/tex]
For the given chemical reaction:
[tex]2Na_2O_2(s)+2H_2O(l)\rightarrow 4NaOH(s)+O_2(g);\Delta H^o_{rxn}=-126kJ[/tex]
By Stoichiometry of the reaction:
If 2 moles of [tex]Na_2O_2[/tex] produces -126 kJ of energy.
Then, 0.128 moles of [tex]Na_2O_2[/tex] will produce = [tex]\frac{-126kJ}{2mol}\times 0.128mol=-8.064kJ[/tex] of energy.
Thus, the enthalpy of the reaction for given amount of [tex]Na_2O_2[/tex] will be -8.064 kJ.
