Question

water is __________ kJ. 2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 25.0 g of Na2O2 with water is __________ kJ. 2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) 40.4 80.8 -126 67.5 20.2

Answer 1

Answer:

20.2 kJ

Explanation:

Based on the information in the reaction, the amount of heat released per mole of Na₂O₂ (the molar enthalpy) is calculated as follows:

126 kJ / 2 mol = 63 kJ/mol Na₂O₂

The number of moles in 25.0g of Na₂O₂ must be calculated using the molecular weight of Na₂O₂ (77.978 g/mol):

(25.0 g)/(77.978 g/mol) = 0.32060 mol Na₂O₂

Thus, the heat released will be:

(63 kJ/mol)(0.32060 mol) = 20.2 kJ

Answer 2

Answer : The amount of heat released by the reaction is, 20.2 kJ

Explanation :

First we have to calculate the number of moles of $Na_2O_2$.

$\text{Moles of }Na_2O_2=\frac{\text{Mass of }Na_2O_2}{\text{Molar mass of }Na_2O_2}$

Molar mass of $Na_2O_2$ = 77.98 g/mole

$\text{Moles of }Na_2O_2=\frac{25.0g}{77.98g/mole}=0.320mole$

Now we have to calculate the heat released during the reaction.

The balanced chemical reaction is:

$2Na_2O_2(s)+2H_2O(l)\rightarrow 4NaOH(s)+O_2(g)$

From the reaction we conclude that,

As, 2 moles of $Na_2O_2$ releases heat = 126 kJ

So, 0.320 moles of $Na_2O_2$ releases heat = $\frac{0.320}{2}\times 126=20.2kJ$

Therefore, the amount of heat released by the reaction is, 20.2 kJ