Question

Suppose 75 metric tons of coal that is 3.0% sulfur by mass is burned at a power plant. During combustion the sulfur is converted into SO2. Antipollution scrubbers installed in the smoke stacks of the power plant capture 3.9 metric tons of this SO2. How efficient are the scrubbers in capturing SO2? How many metric tons of SO2 escaped?

Answer 1

Explanation:

The given data shows that 75 metric tons of coal contains 3.0% sulfur by mass.

Therefore, mass of $SO_{2}$ when 75 metric tons of coal is burned is calculated as follows.

            Mass of sulfur = $\frac{3}{100} \times 75000 kg$

                                     = 2250 kg of S   (as 1 metric tons = 1000 kg)

The equation for conversion will be as follows.

        $S + O_{2} \rightarrow SO_{2}$

Molar mass of sulfur is 32 g/mol and molar mass of $SO_{2}$ is 64 g/mol. That is double the mass of sulfur.

Hence, mass of $SO_{2}$ produced will be as follows.

                   $2 \times 2250 kg$

                   = 4500 kg

As the scrubbers captured 3.9 metric tons or 3900 kg of $SO_{2}$. Therefore, calculate the mass of $SO_{2}$ escaped as follows.

             Mass produced - mass captured

           = 4500 kg - 3900 kg

           = 600 kg

or,       = 0.6 metric tons

Thus, we can conclude that 0.6 metric tons of $SO_{2}$ escaped.