Calculate the enthalpy change in kJ when 21.6g of H2S is burned.

2H2S(g) + 3O2(g) --> 2SO2(g) + 2H20(g) Change in H=-1036kJ/mol

To calculate the enthalpy change (ΔH) when 21.6g of H2S is burned, we need to first determine the number of moles of H2S present.

The molar mass of H2S is:
2(1.008 g/mol) + 32.07 g/mol = 34.076 g/mol

To calculate the number of moles, we will divide the mass of H2S by its molar mass:
Number of moles = Mass (g) / Molar mass (g/mol)
Number of moles = 21.6 g / 34.076 g/mol

Next, we need to find the ΔH for the reaction. As stated in the given equation, the ΔH is -1036 kJ/mol.

Now, since the balanced equation tells us that 2 moles of H2S are required to produce -1036 kJ of energy, we can calculate the enthalpy change for the given mass of H2S.

Enthalpy Change (ΔH) = Number of moles of H2S x ΔH per mole
Enthalpy Change (ΔH) = Number of moles x -1036 kJ/mol

Substituting the number of moles calculated earlier, we can find the enthalpy change:
Enthalpy Change (ΔH) = (21.6 g / 34.076 g/mol) x -1036 kJ/mol

Calculating this expression will give us the enthalpy change in kilojoules (kJ).