2H2S + SO2-> 3S + 2H2O
n (S) = m (S) / M (S) = 16/32 = 0.5 mol
according to the equation of 3 mol of gases (2 mol of hydrogen sulfide and 1 mol of sulfur oxide (IV)), 3 mol of sulfur is obtained, which means 0.5 mol of sulfur will be obtained from 0.5 mol of gas.
the molar volume of gases under normal conditions is 22.4 l / mol, which means that the total volume of the reacted gases is 11.2 l.
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