Question #ca18e

Start by taking a look at the balanced chemical equation

#2"Na"_2"O"_text(2(s]) + 2"H"_2"O"_text((l]) -> 4"NaOH"_text((s]) + "O"_text(2(g])#

You know that the enthalpy change for this reaction is #DeltaH^@ = -"126 kJ"#.

Notice that you have 2 moles of sodium peroxide, #"Na"_2"O"_2#, reacting with 2 moles of water, and producing 4 moles of sodium hydroxide, #"NaOH"#, and oxygen gas.

This tells you that the given enthalpy change of reaction corresponds to the formation of 4 moles of #"NaOH"#. In other words, when 4 moles of #"NaOH"# are formed, #"126 kJ"# of heat are being given off.

An important thing to remember is that if the question asks for heat released, you can skp the minus sign altogether and simply give the value.

SImply put, the expression

the value of the enthalpy change of reaction is - 126 kJ

and the expression

the reaction gives off 126 kJ of heat

are equivalent.

So, if 4 moles will give off #"126 kJ"# of heat, it follows that 2 moles will give off

#2color(red)(cancel(color(black)("moles NaOH"))) * "126 kJ"/(4color(red)(cancel(color(black)("moles NaOH")))) = "63 kJ"#

When 2 moles of sodium hydroxide are formed, the reaction will give off #"63 kJ"# of heat.