What is the maximum mass of S8 that can be produced by combining 80.0 grams of each reactant?
8SO2 + 16H2S -> 3S2 + 16H2O
you have to figure how many moles you have of SO2:80/64
of H2S: 80/34
You need twice as much of H2S as SO2.
If you have more than twices as much, then the limiting reactant is SO2. for S8 (you have S2), it will be then 3/8 of the moles you have of SO2.
Now if you have less than twice as much, then the limiting reactant is H2S, and the moles of S8 will be 3/16 of the amount of H2S you had.
Someone answer the damn question already!
144 g s8
It's actually 113 according to my book.
To determine the maximum mass of S8 that can be produced, you need to consider the stoichiometry, which is the ratio of the reactants and products given in the balanced chemical equation.
Here is the balanced equation:
8SO2 + 16H2S -> 3S2 + 16H2O
From the equation, you can see that the ratio of SO2 to S8 is 8:3. This means that for every 8 moles of SO2, you will produce 3 moles of S8.
First, you need to convert the mass of each reactant to moles using their molar masses:
Molar mass of SO2 = 32.07 g/mol + 2 * 16.00 g/mol = 64.07 g/mol
Molar mass of H2S = 2 * 1.01 g/mol + 32.07 g/mol = 34.09 g/mol
Now, calculate the number of moles for each reactant:
Number of moles of SO2 = 80.0 g / 64.07 g/mol ≈ 1.25 mol
Number of moles of H2S = 80.0 g / 34.09 g/mol ≈ 2.35 mol
Next, you need to determine which reactant limits the formation of S8. To do this, compare the stoichiometric ratio of the reactants. The ratio of SO2 to H2S is 8:16, which can be simplified to 1:2. This means that for every 1 mole of SO2, you need 2 moles of H2S to react completely.
Since you have 1.25 moles of SO2 and 2.35 moles of H2S, the H2S is in excess because you have more moles of H2S than the stoichiometric ratio requires. Therefore, the SO2 is the limiting reactant.
Now, calculate the moles of S8 that can be produced from the limiting reactant using the stoichiometric ratio:
From the balanced equation, 8 moles of SO2 produce 3 moles of S8.
So, 1.25 moles of SO2 will produce (1.25 moles / 8 moles) * 3 moles of S8 ≈ 0.46875 moles of S8.
Finally, convert the moles of S8 to grams using its molar mass:
Molar mass of S8 = 8 * 32.07 g/mol = 256.56 g/mol
Mass of S8 = 0.46875 moles * 256.56 g/mol ≈ 120.04 grams
Therefore, the maximum mass of S8 that can be produced by combining 80.0 grams of each reactant is approximately 120.04 grams.