Question

A certain grade coal contains \(1.6\) per cent sulphur. Assuming that on burning the coal, \(\mathrm{S}\) in it is oxidised to \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{SO}_{2}\) would be formed on burning 1 metric ton \((1000 \mathrm{~kg})\) of coal? (1) 16 (2) \(16000 \times \frac{2}{64}\) (3) \(16000 \times \frac{2}{32}\) (4) \(\frac{16000}{64}\)

Short answer

(3) \frac{16000}{32}.

Step-by-step solution

Calculate Mass of Sulphur

Determine the mass of sulphur in 1 metric ton of coal. Since the coal contains 1.6% sulphur, the mass of sulphur is \[ \text{Mass of } S = \frac{1.6}{100} \times 1000 \text{ kg} = 16 \text{ kg} \]

Convert Mass to Grams

Convert the mass of sulphur from kilograms to grams:\[ 1 \text{ kg} = 1000 \text{ grams} \] Applying this conversion, we get \[ 16 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 16000 \text{ grams} \]

Calculate Moles of Sulphur

Find the number of moles of sulphur using its molar mass (32 g/mol):\[ \text{Moles of } S = \frac{\text{Mass of } S}{\text{Molar Mass}} = \frac{16000 \text{ g}}{32 \text{ g/mol}} = 500 \text{ moles} \]

Determine Moles of SO₂

One mole of sulphur (\text{S}) produces one mole of sulphur dioxide (\text{SO₂}). Therefore, the moles of \text{SO₂} formed are equal to the moles of sulphur:\[ \text{Moles of } SO₂ = 500 \text{ moles} \]

Identify the Matching Option

Compare the calculated value with the given options. The closest equivalent to 500 moles from the given options is \( \frac{16000}{32}=500 \).

Key concepts

Mole Calculation

Mole calculations help us connect the mass of a substance with the number of moles. We use molar mass, which is the weight of one mole of that substance. For example, sulfur (S) has a molar mass of 32 g/mol. To convert mass into moles, we use the formula:

\( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \)

In the exercise, we must first determine the mass of sulfur in the coal. Given that the coal contains 1.6% sulfur, we calculate the mass of sulfur in 1000 kg (1 metric ton) as follows:

\( \text{Mass of } S = \frac{1.6}{100} \times 1000 \text{ kg} = 16 \text{ kg} \)

Next, we convert this mass from kilograms to grams because molar mass is in grams per mole:

\( 16 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 16000 \text{ grams} \)

Finally, we find the number of moles of sulfur:

\( \text{Moles of } S = \frac{16000 \text{ g}}{32 \text{ g/mol}} = 500 \text{ moles} \)

This calculation shows us that 16000 grams of sulfur corresponds to 500 moles of sulfur.

Percent Composition

Percent composition indicates the percentage of each element in a compound by mass. Understanding percent composition is crucial for solving problems related to mixtures, such as coal containing sulfur.

In the given exercise, the coal contains 1.6% sulfur. This information allows us to determine the exact amount of sulfur present by applying the percentage to the total mass of the coal:

\( \frac{1.6}{100} \times 1000 \text{ kg} = 16 \text{ kg} \)

We then convert this mass to grams, crucial for mole calculation:

\( 16 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 16000 \text{ grams} \)

Thus, percent composition helps us convert real-world measurements into useful data for further calculations, like determining the number of moles.

Chemical Reactions

Understanding how chemical reactions work is essential for solving related problems. In the given exercise, sulfur (S) reacts to form sulfur dioxide \( \text{SO}_2 \) when coal burns.

The balanced chemical equation for this reaction is:

\( S + O_2 \rightarrow SO_2 \)

This equation tells us that one mole of sulfur reacts with one mole of oxygen to produce one mole of sulfur dioxide. This 1:1 stoichiometric ratio is key to figuring out how much \( \text{SO}_2 \) forms from a given amount of sulfur.

From the earlier calculation, we found that there are 500 moles of sulfur in 1000 kg of coal. Therefore:

\( \text{Moles of } SO_2 = \text{Moles of } S = 500 \text{ moles} \)

So, 500 moles of sulfur will produce 500 moles of sulfur dioxide. Understanding the stoichiometry of the reaction helps us connect reactants to products effectively.