How many kJ of heat are released by the reaction of 25.0 g of Na2O2(s) in the following reaction? (M = 78.0 g/mol for Na2O2)
1 answer:
-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.
Explanation:
Given:
mass of Na2O2 = 25 grams
atomic mass of Na2O2 = 78 gram/mole
number of mole =
=
=0. 32 moles
The balanced equation for the reaction:
2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ
It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.
similarly 0.3 moles of Na2O2 on reaction would give:
=
x =
= -20.16 KJ
Thus, - 20.16 KJ of energy will be released.
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The enthalpy of fusion for Hg is 2.29 kJ/mol. 0.89 kJ is the energy change when 78.0 g of Hg melts at −38.8°C.
<h3>What is Enthalpy of Fusion ?</h3>The amount of energy needed to change 1 mole of substance under state change at constant temperature and pressure is called enthalpy of fusion. It is also known as Latent heat of fusion. Unit of enthalpy of fusion is kJ/mol.
<h3>How to find the change in energy ? </h3>To find the change in energy use this expression:
q = n ΔH
where
q = Energy change
n = number of moles
ΔH = Molar enthalpy
Number of moles (n) =
=
= 0.39 mol
Now put the values in above formula we get
q = n ΔH
= 0.39 mol × 2.29 kJ/mol
= 0.89 kJ
Thus from the above conclusion we can say that The enthalpy of fusion for Hg is 2.29 kJ/mol. 0.89 kJ is the energy change when 78.0 g of Hg melts at −38.8°C.
Learn more about the Latent heat of fusion here: brainly.com/question/87248
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