# Question 8a001

Apr 27, 2017

The combustion of 130 mg of sulfur creates $\text{1 m"^3}$ of toxic air.

#### Explanation:

The IDLH level of ${\text{SO}}_{2}$ (the level at which it is Immediately Dangerous to Life or Health) is 100 ppm.

∴ In ${\text{1 m}}^{3}$ of air, the toxic volume of ${\text{SO}}_{2}$ is

1 color(red)(cancel(color(black)("m"^3))) × (1000 color(red)(cancel(color(black)("L"))))/(1 color(red)(cancel(color(black)("m"^3)))) × "100 L SO"_2/(10^6 color(red)(cancel(color(black)("L")))) = "0.1 L"

At ambient conditions (say, 20 °C and 1 bar) this corresponds to

n = (pV)/(RT) = (1 color(red)(cancel(color(black)("bar"))) × 0.1 color(red)(cancel(color(black)("L"))))/("0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 293 color(red)(cancel(color(black)("K")))) = "0.0041 mol"#

The equation for the combustion of sulfur is

${\text{S" + "O"_2 → "SO}}_{2}$

Thus,

$\text{Mass of S" = 0.0041 color(red)(cancel(color(black)("mol SO"_2))) × (1 color(red)(cancel(color(black)("mol S"))))/(1 color(red)(cancel(color(black)("mol SO"_2)))) × "32.06 g S"/(1 color(red)(cancel(color(black)("mol S")))) = "0.13 g S" = "130 mg S}$

The combustion of 130 mg of sulfur will create a toxic level of sulfur dioxide in one cubic metre of air.