What is the maximum mass of S8 that can be produced by combining 80g of each reactant?
8SO2 + 16H2S --> 3S8 + 16H2O
i calculated the mass of the reactants
3S8 = 24x 32.066= 769.584 g/mol
16H2 = 32X 1.008= 32.256 g/mol
O= 16X 15.994= 255.904 g/mol
but from there i don't know where to go
To find the maximum mass of S8 that can be produced, we need to determine which reactant is limiting. The limiting reactant is the reactant that is completely consumed in the reaction and limits the amount of product that can be formed.
First, we need to determine the moles of each reactant. We can convert the given masses of reactants (80g each) to moles using their molar masses.
The molar mass of SO2 is 32.066 g/mol, so the moles of SO2 can be calculated as:
moles of SO2 = mass of SO2 / molar mass of SO2 = 80g / 32.066 g/mol
The molar mass of H2S is 34.08 g/mol, so the moles of H2S can be calculated as:
moles of H2S = mass of H2S / molar mass of H2S = 80g / 34.08 g/mol
Now, we can use the balanced equation to determine the ratio of moles of reactants to moles of product. From the balanced equation, we see that 8 moles of SO2 react with 16 moles of H2S to produce 3 moles of S8.
Using the ratio, we can determine the moles of S8 that can be produced if all reactants are completely reacted:
moles of S8 = (moles of H2S * 3) / 16
Finally, we can convert the moles of S8 to grams using the molar mass of S8 (256.52 g/mol):
mass of S8 = moles of S8 * molar mass of S8
This will give us the maximum mass of S8 that can be produced by combining 80g of each reactant.