What is the maximum mass of S8 that can be produced by combining 85.0 g of each reactant?
8SO2 + 16H2S yields 3S8 +16H2O
To determine the maximum mass of S8 that can be produced, we need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed first and determines the amount of product that can be formed.
1. Start by calculating the number of moles of each reactant:
- Moles of SO2 = Mass of SO2 / molar mass of SO2
- Moles of H2S = Mass of H2S / molar mass of H2S
2. Next, calculate the stoichiometric ratio of each reactant by dividing the number of moles of each reactant by their respective stoichiometric coefficients in the balanced equation:
- Stoichiometric ratio of SO2 = Moles of SO2 / coefficient of SO2 in the balanced equation
- Stoichiometric ratio of H2S = Moles of H2S / coefficient of H2S in the balanced equation
3. Identify the limiting reactant. The reactant with the lowest stoichiometric ratio is the limiting reactant.
4. Once the limiting reactant is determined, calculate the theoretical yield of S8 using the stoichiometric ratio of S8 from the balanced equation:
- Moles of S8 = Stoichiometric ratio of S8 × coefficient of S8 in the balanced equation
5. Finally, calculate the mass of S8 by multiplying the number of moles of S8 by the molar mass of S8:
- Mass of S8 = Moles of S8 × molar mass of S8
By following these steps, you will be able to find the maximum mass of S8 that can be produced from the given reactants.
Convert each to mols S8 produced.
mols SO2 = 85/64 = about 1.3 estimated. You need to do it exactly.
mols H2S = 85/34 = about 2.5
Use the coefficients in the balanced equation to convert mols of each to mols S8.
First SO2:
1.3 x (3 mols S8/8 mols SO2) = about 5 mols S8 produced if we had 1.3 mols SO2 and all of the H2S we needed.
Next H2S:
2.5 x (3 mol S8/16 mol H2S) = aboaut 4.5 mol S8 produced if we had 2.5 mol H2S and all of the SO2 we needed.
You have two different values for S; you know one of them must be wrong. The correct value in limiting regent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
mass S8 = mols S8 x molar mass S8 = ?grams.