A coal burning plant that burns 4 metric tons of coal containing 2% sulfur (S) requires 1 metric ton of wet limestone (calcium carbonate) to remove 90% of the sulfur. The process prevents most of the sulfur from being emitted as sulfur dioxide to the air by producing calcium sulfate.

a. show the chemical reaction
b. Calculate percentage of limestone converted to calcium sulfate?

a. The chemical reaction that occurs in the coal burning plant to remove sulfur is as follows:

2 CaCO3 (limestone) + 2 SO2 (sulfur dioxide) + O2 (oxygen) → 2 CaSO4 (calcium sulfate) + 2 CO2 (carbon dioxide)

b. To calculate the percentage of limestone converted to calcium sulfate, we need to consider the stoichiometry of the reaction and the amount of sulfur removed.

Given:
- 4 metric tons of coal containing 2% sulfur
- 1 metric ton of wet limestone is required to remove 90% of the sulfur

First, calculate the amount of sulfur in the coal:
Sulfur in coal = 4 metric tons × 2% = 0.08 metric tons

Since 1 metric ton of wet limestone is required to remove 90% of the sulfur, the amount of limestone needed to remove 0.08 metric tons of sulfur would be:
Limestone needed = 0.08 metric tons / (0.9 × sulfur content in limestone)

Assuming the limestone contains 100% calcium carbonate, the amount of limestone needed would be:
Limestone needed = 0.08 metric tons / (0.9 × 100%) = 0.08 metric tons / 0.9 = 0.0889 metric tons

Now we can calculate the percentage of limestone converted to calcium sulfate:
Percentage of limestone converted = (Weight of calcium sulfate produced / Weight of limestone used) × 100%

Since the stoichiometry of the reaction is 2:2, meaning for every 2 moles of limestone used, 2 moles of calcium sulfate are produced, we can calculate the weight of calcium sulfate produced using the molar masses:

Weight of calcium sulfate produced = (moles of limestone used × molar mass of limestone) × (molar mass of calcium sulfate / molar mass of limestone)

The molar mass of limestone (CaCO3) is 100.09 g/mol, and the molar mass of calcium sulfate (CaSO4) is 136.14 g/mol.

Moles of limestone used = 0.0889 metric tons × (10^6 g /metric ton) / molar mass of limestone

Weight of calcium sulfate produced = (0.0889 metric tons × (10^6 g /metric ton) / molar mass of limestone) × (molar mass of calcium sulfate / molar mass of limestone)

Finally, we can substitute these values into the formula for the percentage of limestone converted:

Percentage of limestone converted = (Weight of calcium sulfate produced / Weight of limestone used) × 100%

a. The chemical reaction involved in this process is as follows:

2 CaCO3 + 2 SO2 + O2 -> 2 CaSO4 + 2 CO2

This reaction shows that two molecules of calcium carbonate (CaCO3) react with two molecules of sulfur dioxide (SO2) and oxygen (O2) to produce two molecules of calcium sulfate (CaSO4) and two molecules of carbon dioxide (CO2).

b. To calculate the percentage of limestone converted to calcium sulfate, we first need to determine the amount of sulfur (in metric tons) that the limestone can remove. We know that 4 metric tons of coal contains 2% sulfur, so:

Amount of sulfur = 4 metric tons of coal * 2% = 0.08 metric tons of sulfur

Now, we can calculate the amount of limestone required to remove 90% of the sulfur:

Limestone required = 0.08 metric tons of sulfur / 90% = 0.0889 metric tons

Since the process requires 1 metric ton of wet limestone, we can calculate the percentage conversion as follows:

Percentage conversion = (0.0889 metric tons / 1 metric ton) * 100%
= 8.89%

Therefore, approximately 8.89% of the limestone is converted to calcium sulfate during this process.