Question

Can someone explain how to do this? Consider the following equation: 2ZnS(s) + 3O2(g) ---> 2ZnO(s) + 2SO2(g), ΔH° = –878.2 kJ. What is the enthalpy change when 223.9 g ZnS(s) reacts in excess oxygen?

Answer 1

In this question a chemical equation is given;

2ZnS + 3`O_2` -----------> 2 ZnO + 2S`O_2` Enthalpy Change = -878.2 kJ

as per the equation 2 moles of ZnS releases -872.8kJ of energy on reaction with Oxygen.

2 moles of ZnS = -872 kJ of heat energy

In terms of mass, 2 moles of ZnS has mass 2 x 97.5 g = 195 g

195 g of ZnS on reaction gives -878.2 kJ of heat energy.

1 g of ZnS releases = -878.2 kJ / 195 of heat energy

223.9 g of ZnS releases = -878.2 kJ x 223.9g / 195g

= -1008 kJ of heat energy.