2H2S + 3O2 = 2H2O + 2SO2
H2S is a limiting reagent and 2 moles of it will burn out in 3 moles of O2. 3 moles of O2 is 6 dm3, which means 4 dm3 of oxygen will remain in products as an extra.
By stoichiometry, 2 moles of each product was produced, that means 4 dm3 H2O and 4 dm3 of SO2 were produced.
Final volume = 4 dm3 (O2 not used) + 4dm3 (SO2) + 4 dm3 (H2O) = 12 dm3