In coal, pyrites `(FeS_(2))` is present as a pollution-causing impurity, which is removed by combustion.
`2FeS_(2) + 5O_(2) rarr 4SO_(2) + 2FeO`.
A process designed to remove orgainc sulphur from coal prior to combustion involves the reaction.
`X - S - Y + 2NaOH rarr X - O - Y + Na_(2)S + H_(2) O`
`CaCO_(3) rarr CaO + CO_(2)`
`Na_(2)S + CO_(2) + H_(2)O rarr Na_(2) CO_(3) + H_(2) S`
`CaO + H_(2)O rarr Ca (OH)_(2)`
`Na_(2) CO_(3) + Ca(OH)_(2) rarr CaCO_(3) + 2NaOH`
In the processing in 320 metric tons of a coal having 1.0% sulphur content, how much limestone `(CaCO_(3))` must be edecomposed to provied enough `Ca (OH)_(2)` to regenerate the `NaOH` used in the original leaching step?

Options

A. 2.0 metric ton
B. 4.0 metric ton
C. 8.0 metric ton
D. 10.0 metric ton

Explanation

To solve the problem of how much limestone `(CaCO₃)` must be decomposed to provide enough `Ca(OH)₂` to regenerate the `NaOH` used in the leaching step, we will follow these steps:

### Step 1: Calculate the amount of sulfur in coal
Given that the coal weighs 320 metric tons and has a sulfur content of 1.0%, we first convert the weight of coal into grams:
\[ 320 \text{ metric tons} = 320 \times 10^6 \text{ grams} \]
Next, we calculate the amount of sulfur:
\[ \text{Weight of sulfur} = \frac{1}{100} \times 320 \times 10^6 = 3.2 \times 10^6 \text{ grams} \]

### Step 2: Calculate the number of moles of sulfur
Using the molar mass of sulfur (S), which is 32 g/mol, we can find the number of moles of sulfur:
\[ \text{Number of moles of sulfur} = \frac{\text{Weight of sulfur}}{\text{Molar mass of sulfur}} = \frac{3.2 \times 10^6 \text{ grams}}{32 \text{ g/mol}} = 10^5 \text{ moles} \]

### Step 3: Relate moles of sulfur to moles of calcium carbonate
From the reaction equations, we know that:
1 mole of sulfur reacts with 1 mole of calcium carbonate `(CaCO₃)`. Therefore, the number of moles of `CaCO₃` required is also:
\[ \text{Number of moles of } CaCO₃ = 10^5 \text{ moles} \]

### Step 4: Calculate the weight of calcium carbonate
The molar mass of calcium carbonate `(CaCO₃)` is 100 g/mol. To find the weight of `CaCO₃` required:
\[ \text{Weight of } CaCO₃ = \text{Number of moles} \times \text{Molar mass} = 10^5 \text{ moles} \times 100 \text{ g/mol} = 10^7 \text{ grams} \]

### Step 5: Convert grams to metric tons
Since 1 metric ton = \(10^6\) grams, we convert grams to metric tons:
\[ \text{Weight of } CaCO₃ = \frac{10^7 \text{ grams}}{10^6 \text{ grams/metric ton}} = 10 \text{ metric tons} \]

### Final Answer
Thus, the amount of limestone `(CaCO₃)` that must be decomposed is **10 metric tons**.

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