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

2 answers

2Na + 2H2O ==> 2NaOH + H2

368 kJ x (0.113 mol/2 mol) = ?
2Na + 2H2O → H2 + 2NaOH is the balanced equation. from that, .113 mole/2=.0565 x -368 = -20.79kj or 20.79 kj heat is created in the reaction.