What is the maximum mass of S8 that can be produced by combining 79.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O

Answer 1

This is a problem mainly dealing with molar masses, and converting between grams and moles.

First, lets establish what the molar masses of the chemicals we care about are:

SO2= 64.06 g/mol

H2S= 34.08 g/mol

S8= 256.53 g/mol

Now that we have those, we can convert the gram amounts of the reactants into moles

79g SO2/64.06 g/mol= 1.23 mol. SO2

79g H2S/34.08 g/mol= 2.32 mol. H2S

Now we have the molar amounts of each reactant. We have already been given the properly balanced stoichometric reaction, and thus know that the molar ratio of SO2 consumed to H2S in the reaction is 8 to 16. This signifies for every 8 moles of SO2, 16 moles of H2S are required to produce a subsequent 3 moles of S8 and 16 moles of H2O.

In this reaction, the limiting reagent is H2S, as one would need approximately 2.46 moles of H2S to fully consume the available SO2.

Now that we know the limiting reagent, and based on the stoichiometric coefficients, we know that for every 16 moles of H2S consumed, 3 moles of S8 are produced, and can use that to calculate the amount of S8 produced in moles.

2.32 mol. H2S/ ? mol. S8 = 16 mol. H2S/ 3 mol. S8

(2.32)*(3)= (16)*(?)

(2.32*3)/16=?

? mol. S8 = 0.435

We have now solved for the moles of S8 produced by the reaction, and can convert the moles to grams using the molar mass listed above to come to the final answer of the maximum mass of S8 produced by this reaction.

0.435 mol. S8 * 256.53 g/mol =111.59

Final Answer: 111.59 grams is the maximum mass of S8 produced by this reaction