Calculate the value of q (kj) in this exothermic reaction when 5.40 g of hydrogen peroxide decomposes at constant pressure?

Hydrogen peroxide decomposes to yield water and oxygen gas 
That is; H2O2 (l) = H2O (l) + O2(g)
The standard heat of formation; H2O2 (l) = -187.6 kJ/mol; H2O(l) = -285.8 kJ/mol
1 mole of hydrogen peroxide contains 34 g
Thus, 5.4 g contains 5.4/34 = 0.1588 moles 
The moles of water produced will also be equivalent to 0.1588 moles 
Heat = heat of formation of product - reactant
Therefore; Heat = (0.1588 moles × -285,8 )- (0.1588× -187.6)
                           =  -15.594 kJ

Which metal is less dense