Question

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?

Answer 1

`"7.14 g"`

Explanation:

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!