Question

The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 10.0 g of Na2O2 with water is ________ kJ. 2Na2O2 (s) + 2H2O (l) → 4NaOH (s) + O2 (g)

Answer 1

Answer: The enthalpy of the reaction for given amount of $Na_2O_2$ will be -8.064 kJ.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Given mass of $Na_2O_2$ = 10 g

Molar mass of $Na_2O_2$ = 78 g/mol

Putting values in above equation, we get:

$\text{Moles of }Na_2O_2=\frac{10g}{78g/mol}=0.128mol$

For the given chemical reaction:

$2Na_2O_2(s)+2H_2O(l)\rightarrow 4NaOH(s)+O_2(g);\Delta H^o_{rxn}=-126kJ$

By Stoichiometry of the reaction:

If 2 moles of $Na_2O_2$ produces -126 kJ of energy.

Then, 0.128 moles of $Na_2O_2$ will produce = $\frac{-126kJ}{2mol}\times 0.128mol=-8.064kJ$ of energy.

Thus, the enthalpy of the reaction for given amount of $Na_2O_2$ will be -8.064 kJ.