# 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?

Nov 20, 2017

$\text{7.14 g}$

#### Explanation:

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

${\text{S"_8(s) + 8"O"_2(g) -> 8"SO}}_{2} \left(g\right)$

Okay, now we can use the ideal gas law:

$P V = n R T$

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 ${\text{S}}_{8}$ and ${\text{SO}}_{2}$. First, we must find the number of moles of S8. Let's use the ideal gas law:

$P = 1.00 a t m$

$V = 1 L$

$R = 0.0821 \frac{L \cdot a t m}{m o l \cdot K}$

T = 600 ˚ C = 873 K

$\left(1.00 a t m\right) \left(1 L\right) = n \left(0.0821 \frac{L \cdot a t m}{m o l \cdot K}\right) \left(873 K\right)$

$n = 0.014$ mols

Now, we can use stoichiometry to find the number of moles of ${\text{SO}}_{2}$. From our balanced equation, we find that, for every mole of ${\text{S}}_{8}$, we get $8$ moles of ${\text{SO}}_{2}$:

("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 ${\text{SO}}_{2}$ is:

$32 \frac{g}{m o l} + 16 \frac{g}{m o l} \cdot 2 = 64 \frac{g}{m o l}$

Finally, using the number of moles and the molar mass, we can determine the mass of ${\text{SO}}_{2}$ produced:

$0.11 m o l \cdot 64 \frac{g}{m o l} = 7.14 g$

Hope this helps!