Respuesta :
Answer : The amount of heat released by the reaction is, 20.2 kJ
Explanation :
First we have to calculate the number of moles of [tex]Na_2O_2[/tex].
[tex]\text{Moles of }Na_2O_2=\frac{\text{Mass of }Na_2O_2}{\text{Molar mass of }Na_2O_2}[/tex]
Molar mass of [tex]Na_2O_2[/tex] = 77.98 g/mole
[tex]\text{Moles of }Na_2O_2=\frac{25.0g}{77.98g/mole}=0.320mole[/tex]
Now we have to calculate the heat released during the reaction.
The balanced chemical reaction is:
[tex]2Na_2O_2(s)+2H_2O(l)\rightarrow 4NaOH(s)+O_2(g)[/tex]
From the reaction we conclude that,
As, 2 moles of [tex]Na_2O_2[/tex] releases heat = 126 kJ
So, 0.320 moles of [tex]Na_2O_2[/tex] releases heat = [tex]\frac{0.320}{2}\times 126=20.2kJ[/tex]
Therefore, the amount of heat released by the reaction is, 20.2 kJ
The amount of heat that is released by the chemical reaction of 25.0 g of [tex]Na_{2}O_2[/tex] with water is -20.223 Joules.
Given the following data:
- Mass of [tex]Na_{2}O_2[/tex] = 25.0 grams
- Enthalpy of combustion = -126 kJ/mol
To find the amount of heat that is released by the chemical reaction of 25.0 g of [tex]Na_{2}O_2[/tex] with water:
First of all, we would determine the number of moles of [tex]Na_{2}O_2[/tex] in this chemical reaction:
[tex]2Na_2O_2_{(s)} + 2H_2O_{(l)}[/tex] ------> [tex]4NaOH_{(s)} + O_2_{(g)}[/tex]
[tex]Number\;of\;moles \;(Na_{2}O_2)= \frac{Mass\; of\;Na_{2}O_2}{Molar\;mass\;of\;Na_{2}O_2}[/tex]
Substituting the values into the formula, we have;
[tex]Number\;of\;moles \;(Na_{2}O_2)= \frac{25.0}{77.98}[/tex]
Number of moles ([tex]Na_{2}O_2[/tex]) = 0.321 moles.
Now, we can find the quantity of heat released when [tex]Na_{2}O_2[/tex] reacts with water:
2 mole of [tex]Na_{2}O_2[/tex] = -126 kJ/mol
0.321 mole of [tex]Na_{2}O_2[/tex] = X kJ/mol
Cross-multiplying, we have:
[tex]2X = 0.321[/tex] × [tex](-126)[/tex]
[tex]2X = -40.446\\\\X = \frac{-40.446}{2}[/tex]
X = -20.223 Joules.
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