Question

The reaction CaO(s)+ H2O(l)—-> ca(OH)2 delta H= -65.2 kJ

Answer 1

Answer:

There is 582 kJ of energy released. Δ H rxn  for 500g=  −  582 kJ  ( negative because the energy is released).

Explanation:

Step 1: The balanced equation

CaO(s)+ H2O(l) → ca(OH)2     ΔH= -65.2 kJ

In this situation, 1 mole of CaO consumed, needs 1 mole of H2O to produce 1 mole of Ca(OH)2

The enthalpy change of this reaction is  Δ H rxn =  −  65.2 kJ

This means the reaction gives off  65.2 kJ  of heat when 1 mole of CaO and 1 mole of H2O react.

Step 2: Calculate number of moles in 500 g of CaO

Number of moles = mass of CaO / Molar mass of CaO

Number of moles = 500 grams / 56.077 g/mol = 8.92 moles

Step 3: Calculate energy released

Since there is released 65.2 kJ when 1 mole of CaO reacts; for 8.92 moles of CaO there will be released 8.92 * 65.2 kJ = 581.584 ≈ 582 kJ

.

There is given off 582 kJ of energy

Δ H rxn  for 500g=  −  582 kJ