The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is s(s) + o2(g) → so2(g) how much sulfur (in tons), present in the original materials, would result in that quantity of so2?

Mass of sulfur dioxide = 26 million tons = [tex]235.872\times 10^{11}g[/tex] (Conversion factor: [tex]\text{1 million ton}=9.072\times 10^{11}g[/tex] )

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)

Given mass of sulfur dioxide = [tex]235.872\times 10^{11}g[/tex]

Molar mass of sulfur dioxide = 64 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of sulfur dioxide}=\frac{235.872\times 10^{11}g}{64g/mol}=3.70\times 10^{11}mol[/tex]

For the given chemical equation:

[tex]S(s)+O_2(g)\rightarrow SO_2(g)[/tex]

By Stoichiometry of the reaction:

1 mole of sulfur dioxide is produced by 1 mole of Sulfur.

So, [tex]3.70\times 10^{11}[/tex] moles of sulfur dioxide will be produced by = [tex]\frac{1}{1}\times 3.70\times 10^{11}=3.70\times 10^{11}[/tex] moles of sulfur.

Calculating the mass of sulfur by using equation 1, we get:

Molar mass of sulfur = 32 g/mol

Moles of sulfur = [tex]3.70\times 10^{11}mol[/tex]

Putting values in equation 1, we get:

[tex]3.70\times 10^{11}mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(32g/mol\times 3.70\times 10^{11}mol)=118.4\times 10^{11}g[/tex]

Converting this into tons, we use the conversion factor:

1 ton = 907185 grams

So, [tex]118.4\times 10^{11}g\times \frac{1ton}{907185g}=1.305\times 10^{-4}tons[/tex]

Hence, the mass of sulfur produced is [tex]1.305\times 10^{-4}tons[/tex]