How much heat is liberated when 0.113 mole of sodium
reacts with excess water according to the following equation?
DH = -368 kJ/mol rxn
See your first post. It has two answers (that are the same).
i don't know, too
To calculate the amount of heat liberated when 0.113 mole of sodium reacts with excess water, we need to use the given enthalpy change value (DH) of the reaction. The DH value (-368 kJ/mol rxn) represents the enthalpy change for one mole of the reaction.
Since we want to find the heat liberated for 0.113 mole of sodium, we can set up a proportion using the moles of sodium as a ratio:
DH for 0.113 mol of sodium / DH for 1 mol of the reaction = Heat liberated for 0.113 mol of sodium / Heat liberated for 1 mol of the reaction
Plugging in the values we know:
Heat liberated for 0.113 mol of sodium = (DH for 0.113 mol of sodium / DH for 1 mol of the reaction) * Heat liberated for 1 mol of the reaction
Heat liberated for 0.113 mol of sodium = (0.113 mol / 1 mol) * (-368 kJ/mol rxn)
Now we can calculate the heat liberated for 0.113 mole of sodium:
Heat liberated for 0.113 mol of sodium = (0.113) * (-368 kJ/mol rxn)
Heat liberated for 0.113 mol of sodium = -41.584 kJ
Therefore, when 0.113 mole of sodium reacts with excess water, approximately -41.584 kJ of heat is liberated.