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What is the maximum mass of S8 that can be produced by combining 77.0 g of each reactant?
8SO2+16H2S⟶3S8+16H2O
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♦ Relevant knowledge
The stoichiometry of mass-to-mass is commonly used to quantify the quantity of products or reactants that make up an chemical reaction. In some instances the quantity of the reactant isn’t enough to make a certain product. This problem is usually solved by determining the limiting reaction which is the one which is used completely during this chemical process. The amount of moles of the reaction that is limited be used later to determine the amount of compounds that result from the reaction.
This is the result.
8SO2 + 16H2S
3S8 + 16H2O
Let’s find the moles reactants
Molar mass of SO2 = 64.06g/mol
Molar mass H2S = 34.08g/mol
using formula Moles = mass / molar mass
Moles of SO2 = 77.0g / 64.06g/mol = 1.2020mmol SO2
Moles of H2S = 77.0g/34.08g/mol = 2.2594mol H2S
Let’s now see how H2S reacts to moles of SO2
Then, 2.2594 mo H2S would react totally with 2.2594 mo ( 3 mol S8 /16 mol H2S) = 0.42 Mol SO2
However, as per the above calculation, there is more SO2 available
So, SO2 is excessive and H2S the limiting reactant
Thus,
Maximum mass of S8 = 2.2594 mole H2S
(3.mol S8/16.mol H2S) 
Molar mass of S8
= 2.2594 mo H2S
(3.mol S8/16 mol H2S). 
256.52 g/mol
= 108.67g S8
Our answer is 108.67 g S8