# If the coal is burned in a power plant that uses 2000 tons of coal per day, what mass of calcium oxide is required daily to eliminate the sulfur dioxide? How many grams of calcium sulfite are produced daily by this power plant?

## A particular coal contains 2.5% sulfur by mass. When this coal is burned at a power plant, the sulfur is converted into sulfur dioxide gas, which is a pollutant. To reduce sulfur dioxide emissions, calcium oxide (lime) is used. The sulfur dioxide acts with calcium oxide to form solid calcium sulfite

Sep 3, 2016

Mass of $C a S {O}_{3}$ produced daily$=$ $1.70 \times {10}^{8} g$

#### Explanation:

$S {O}_{2} \left(g\right) + C a O \left(s\right) \rightarrow C a S {O}_{3} \left(s\right)$

$\text{1 US ton}$ $=$ $907.19 \cdot k g$

$\text{Mass of coal(kg)}$ $=$ $2000 \cdot {\text{ton"xx907*kg*"ton}}^{-} 1$ $=$ $1.814 \times {10}^{6} \cdot k g$.

$\text{Mass of sulfur}$ $=$ 1.814xx10^6*kgxx2.5% $=$ $45 , 350 \cdot k g$

$\text{Moles of sulfur}$ $=$ $\frac{45 , 350 \cdot k g \times {10}^{3} \cdot g \cdot k {g}^{-} 1}{32.06 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.41 \times {10}^{6} \cdot m o l$.

And thus we may have $1.41 \times {10}^{6} \cdot m o l$ $S {O}_{2}$ produced during the daily operation.

Given the equivalence of sulfur dioxide and calcium oxide in the first equation, clearly we need $1.41 \times {10}^{6} \cdot m o l$ $C a O$ to produce $1.41 \times {10}^{6} \cdot m o l$ $C a S {O}_{3}$.

$\text{Mass of calcium sulfite produced daily}$ $=$ $1.41 \times {10}^{6} \cdot m o l \times 120.17 \cdot g \cdot m o {l}^{-} 1$ $=$ $1.70 \times {10}^{8} \cdot g$ $=$ $1.70 \times {10}^{5} \cdot k g$.

There is a lot of arithmetic here; please go over my calculations; there is no money back guarantee.