Question

A boiler burns coal at a rate of 1 kg/s. If the coal has 3% sulfur content, assuming that there is no sulfur in ash, SO2 emitted is ________ kg/day. (rounded off to nearest integer)

Hint:

- The conversion factor for sulfur to SO2 is 2 (because SO2 has twice the molecular weight of sulfur).

Answer 1

Step 1: Understanding the given data.
Coal burn rate = 1 kg/s
Sulfur content in coal = 3%
The amount of sulfur in the coal is \( 1 \, \text{kg/s} \times 0.03 = 0.03 \, \text{kg/s} \).
Step 2: Convert sulfur to SO2.
For each kg of sulfur in the coal, the equivalent mass of SO2 is given by the molecular weight ratio of SO2 to sulfur. The molecular weight of sulfur (S) is 32 g/mol, and for SO2, it is 64 g/mol. Hence, the ratio is: \[ \frac{64}{32} = 2. \] Therefore, for every kg of sulfur, 2 kg of SO2 is emitted.
Step 3: Calculate the SO2 emission rate.
The rate of SO2 emission is: \[ \text{SO2 emission rate} = 0.03 \, \text{kg/s} \times 2 = 0.06 \, \text{kg/s}. \]
Step 4: Convert SO2 emission to kg/day.
To find the total emission in a day: \[ \text{SO2 emission per day} = 0.06 \, \text{kg/s} \times 86400 \, \text{seconds/day} = 5184 \, \text{kg/day}. \]
Step 5: Final answer.
The SO2 emitted is \( \boxed{5184} \, \text{kg/day}. \)