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?

Question

Serenity Allen · Accepted Answer

#"7.14 g"#

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!

Ellie Andrews · Answer

undefined To find the mass of SO2 gas obtained when 1 liter of S8(g) is burned in excess pure O2, you'll first need to determine the stoichiometry of the reaction between S8 and O2 to produce SO2. The balanced chemical equation is:

S8(g) + 8 O2(g) -> 8 SO2(g)

From the equation, you can see that 1 mole of S8 reacts with 8 moles of O2 to produce 8 moles of SO2.

Next, you need to calculate the number of moles of S8 present in 1 liter of S8 vapor at 600 ˚C and 1.00 atm using the ideal gas law:

PV = nRT

where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

Given: 
P = 1.00 atm
V = 1.00 L
T = 600 ˚C + 273.15 = 873.15 K

Now, solve for n:

n = (P * V) / (R * T)

Substitute the given values:

n = (1.00 atm * 1.00 L) / (0.0821 L·atm/mol·K * 873.15 K)
  ≈ 0.0115 moles of S8

Now, using the stoichiometry of the reaction, you can calculate the moles of SO2 produced:

moles of SO2 = moles of S8 * (moles of SO2 / moles of S8)

moles of SO2 = 0.0115 moles * (8 moles SO2 / 1 mole S8)
             ≈ 0.092 moles of SO2

Finally, use the molar mass of SO2 (approximately 64 grams per mole) to find the mass of SO2 produced:

mass of SO2 = moles of SO2 * molar mass of SO2
            ≈ 0.092 moles * 64 g/mol
            ≈ 5.89 grams of SO2.