The interaction of hydrogen sulfide with a lack of oxygen formed 16 g of sulfur. The total volume of reacted gases is

with a lack of oxygen there is a process
2H2S + O2-> 2S + 2H2O
n (S) = m (S) / M (S) = 16/32 = 0.5 mol
after the equation, 2 moles of sulfur is obtained from 3 moles of gases (2 moles of hydrogen sulfide and 1 mole of oxygen),
then 0.5 mol of sulfur will be obtained from 0.75 mol of gases.
the molar volume of gases under normal conditions – 22.4 l / mol means
the total volume of reacted gases is 22.4 * 0.75 = 16.8 liters.