Calculating SO₂ from Organic Waste

We need to determine the mass of sulfur dioxide ($SO_2$) produced from burning 1 kg of the organic waste represented by the formula $C_{240}O_{200}H_{180}N_5S$. We are given the atomic weights: S=32, H=1, C=12, O=16, N=14.

Molecular Weight of Organic Waste

First, calculate the molecular weight (MW) of the organic waste using the given formula and atomic weights:

$MW_{waste} = (240 \times MW_C) + (200 \times MW_O) + (180 \times MW_H) + (5 \times MW_N) + (1 \times MW_S)$

$MW_{waste} = (240 \times 12) + (200 \times 16) + (180 \times 1) + (5 \times 14) + (1 \times 32)$

$MW_{waste} = 2880 + 3200 + 180 + 70 + 32 = 6362 \text{ g/mol}$

Sulfur Mass Calculation

Next, find the mass fraction of Sulfur (S) in the waste:

$\text{Mass fraction of S} = \frac{\text{Atomic weight of S}}{MW_{waste}} = \frac{32}{6362}$

Now, calculate the mass of Sulfur present in 1 kg (1000 g) of the waste:

$\text{Mass of S in 1000 g waste} = 1000 \text{ g} \times \frac{32}{6362} \approx 5.02986 \text{ g}$

SO₂ Generation Calculation

The problem states that all Sulfur (S) is completely converted to Sulfur Dioxide ($SO_2$) upon burning. The reaction is effectively $S \rightarrow SO_2$. This means 1 mole of S produces 1 mole of $SO_2$. We use the molar masses to convert the mass of S to the mass of $SO_2$.

• Molar mass of S = 32 g/mol
• Molar mass of $SO_2 = MW_S + (2 \times MW_O) = 32 + (2 \times 16) = 32 + 32 = 64$ g/mol

The mass ratio of $SO_2$ to S is $64/32 = 2$. Therefore, the mass of $SO_2$ produced is twice the mass of S burned.

$\text{Mass of } SO_2 \text{ produced} = (\text{Mass of S}) \times \frac{MW_{SO_2}}{MW_S}$

$\text{Mass of } SO_2 \text{ produced} = 5.02986 \text{ g} \times \frac{64}{32} = 5.02986 \text{ g} \times 2 \approx 10.05972 \text{ g}$

Final Answer Formatting

Rounding the result to 1 decimal place, we get 10.1 g.

The calculated value of 10.1 g lies within the provided correct answer range of 9.9 to 10.2.