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?

1 Answer
Nov 20, 2017

Answer:

#"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!