One liter of sulfur vapor, S8(g) , at 600 ˚ C and 1.00 atm is burned in excess pure O2 to give SO2, measured at the same temperature and pressure. What mass of SO2 gas is obtained?

Don't we all love stoichiometry? The best way to start these types of problems is with a balanced equation:

`"S"_8(s) + 8"O"_2(g) -> 8"SO"_2(g)`

Okay, now we can use the ideal gas law:

`PV = nRT`

where `P` is pressure, `V` is volume, `n` is number of moles, `R` is the gas constant, and `T` is the temperature.

Since we have excess (unlimited) oxygen gas, we can ignore that for this problem. Let's just focus on `"S"_8` and `"SO"_2`. First, we must find the number of moles of S8. Let's use the ideal gas law:

`P = 1.00 atm`

`V = 1 L`

`R = 0.0821 (L*atm)/(mol*K)`

`T = 600 ˚ C = 873 K`

`(1.00 atm)(1 L) = n(0.0821 (L*atm)/(mol*K))(873 K)`

`n = 0.014` mols

Now, we can use stoichiometry to find the number of moles of `"SO"_2`. From our balanced equation, we find that, for every mole of `"S"_8`, we get `8` moles of `"SO"_2`:

`("0.014 mol S"_8) xx ("8 mol SO"_2)/("1 mol S"_8) = "0.11 moles SO"_2`

Using the periodic table, we can determine that the molar mass of `"SO"_2` is:

`32 g/(mol) +16 g/(mol) *2 = 64 g/(mol)`

Finally, using the number of moles and the molar mass, we can determine the mass of `"SO"_2` produced:

`0.11 mol * 64 g/(mol) = 7.14 g`

Hope this helps!