What is the maximum mass of S8 that can be produced by combining 85.0 g of each reactant?
8SO2 + 16H2S yields 3S8 +16H2O
I don't believe Junday into account the 85 grams of each reactant.
mols SO2 = 85/64 = about about 1.3
mols H2S = 85/34 = about about 2.5
Calculate LR from that and I believe the LR is H2S. Check it out.
To find the maximum mass of S8 that can be produced, you need to determine which of the reactants limits the formation of S8. This can be done by finding the number of moles of S8 produced from each reactant and comparing them.
First, let's calculate the number of moles of each reactant:
1. Calculate the number of moles of SO2:
- Given mass of SO2: 85.0 g
- Molar mass of SO2: 64.07 g/mol
- Number of moles of SO2 = (mass of SO2) / (molar mass of SO2)
2. Calculate the number of moles of H2S:
- Given mass of H2S: 85.0 g
- Molar mass of H2S: 34.08 g/mol
- Number of moles of H2S = (mass of H2S) / (molar mass of H2S)
Now, we need to find the number of moles of S8 produced from each reactant. From the balanced equation:
- 8 moles of SO2 will produce 3 moles of S8.
- 16 moles of H2S will produce 3 moles of S8.
3. Calculate the number of moles of S8 produced from SO2:
- Number of moles of SO2 obtained earlier
- Number of moles of S8 from SO2 = (Number of moles of SO2) x (3 moles of S8 / 8 moles of SO2)
4. Calculate the number of moles of S8 produced from H2S:
- Number of moles of H2S obtained earlier
- Number of moles of S8 from H2S = (Number of moles of H2S) x (3 moles of S8 / 16 moles of H2S)
Now, let's determine which reactant limits the formation of S8. The reactant that produces the smallest amount of S8 will be the limiting reactant.
Compare the number of moles of S8 produced from SO2 and H2S. If the number of moles of S8 from SO2 is smaller, then SO2 is the limiting reactant. Otherwise, if the number of moles of S8 from H2S is smaller, then H2S is the limiting reactant.
Once you have identified the limiting reactant, use its molar ratio with S8 to calculate the maximum mass of S8 that can be produced. Multiply the number of moles of the limiting reactant by the molar mass of S8 to get the maximum mass.
5. Calculate the maximum mass of S8:
- If SO2 is the limiting reactant:
- Calculate the maximum number of moles of S8 from SO2 by multiplying the number of moles of SO2 produced earlier by (3 moles of S8 / 8 moles of SO2).
- Maximum mass of S8 from SO2 = (maximum number of moles of S8 from SO2) x (molar mass of S8)
- If H2S is the limiting reactant:
- Calculate the maximum number of moles of S8 from H2S by multiplying the number of moles of H2S produced earlier by (3 moles of S8 / 16 moles of H2S).
- Maximum mass of S8 from H2S = (maximum number of moles of S8 from H2S) x (molar mass of S8)
Compare the maximum masses of S8 obtained from both reactants, and the smaller mass will be the maximum mass of S8 that can be produced by combining 85.0 g of each reactant.
It's a limiting reagent problem.
8mol*(64.07 g/mol)=512.56 g
16mol*(34.07 g/mol)=545.12 g
Therefore, SO2 is the limiting reagent. Figure out the rest