What is the maximum mass of S8 in grams that can be produced by combining 87.1 g of each reagent? Do not include units in your answer. Answer to three significant figures.
8SO2 + 16H2S → 3S8 + 16H2O
Since the required amounts of the two reagents have almost the same mass, H2S will be the limiting one, since twice as many moles of that are required.
So, it will be the limiting reagent.
So, figure how many moles of H2S are in 87.1g, and calculate the mass of 3/16 that many moles of S8.