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