The value of delta H for the reaction of 2Na2O2(s) + 2H20(l)----> 4NaOH(s)+O2(g) is -126KJ When 2.0 moles of NaOH is formed?

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To find the heat exchanged (ΔH) when a certain amount of product is formed, we need to use the concept of stoichiometry and the given ΔH value for the reaction.

First, let's determine the moles of NaOH formed using its given coefficient in the balanced chemical equation:

2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g)

From the equation, we can see that for every 2 moles of Na2O2, 4 moles of NaOH are formed. Therefore, the ratio of NaOH to Na2O2 is 4:2, which simplifies to a ratio of 2:1.

Since we want to find the ΔH when 2.0 moles of NaOH is formed, we can conclude that 1 mole of Na2O2 has reacted. This means that the reaction has released -126 kJ of heat energy.

So, when 2.0 moles (twice as much) of NaOH is formed, we still have the same amount of Na2O2 reacting, meaning that the heat change remains the same. Therefore, the value of ΔH is still -126 kJ.

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