Question

Coal can undergo an incomplete combustion in the absence of a plentiful supply of air to produce deadly carbon monoxide gas. What volume of carbon monoxide is produced at SATP by the incomplete combustion of 150. kg of coal?

Answer 1

SATP conditions:

Standard ambient temperature is 25 ℃ or 298.15 K or 77 ℉.

Standard pressure is 1 bar or 100.000 kPa or 750.06 mmHg.

150. kg × (1000 g / 1 kg) = 150000 g

Solution:

Coal = C (carbon)

The balanced chemical equation:

2C + O₂ = 2CO

According to the chemical equation: n(C) = n(CO)

Moles of C = n(C) = Mass of C / Molar mass of C

The molar mass of C is 12 g/mol.

n(C) = (150000 g) / (12 g/mol) = 12500 mol

n(CO) = n(C) = 12500 mol

The Ideal Gas Law is: PV=nRT

We can rearrange the Ideal Gas Law to get:

V = nRT/P

At SATP,

V(CO) = (12500 mol × 0.08314 L bar K^-1 mol^-1 × 298.15 K) / (1 bar) = 309852.3875 L = 309852.4 L

Answer: 309852.4 L of carbon monoxide (CO) is produced at SATP.