Question

Propane is a common fuel for rural buildings, and assuming the connection between the tank and burner facilitates sufficient heat transfer for propane vaporization, a typical household water heater burns propane gas (C₃​H₈​(g)) with 20% excess air to ensure complete combustion. The fuel and dry air enter at standard pressure and temperature (1 atm, 298 K). The exhaust gases are kept at a high temperature (450 K) to prevent condensation and corrosion in the heat exchanger and the metal exhaust stack. The steady-state heat transfer rate from the combustion products to the water is 50,000 kJ/hr. Find the air/fuel ratio at p.

Answer 1

The air/fuel ratio is 5 moles of oxygen per mole of propane.

Step-by-step explanation:

First, let's balance the equation for the complete combustion of propane gas:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

The equation shows that 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide and 4 moles of water.

To find the air/fuel ratio, we need to calculate the moles of propane and the moles of oxygen.

Since we are assuming 20% excess air, the moles of propane will be the limiting reactant. Using the molar mass of propane (44.1 g/mol) and the mass of propane (25.0 g), we can calculate the moles of propane:

Moles of propane = Mass of propane / Molar mass of propane = 25.0 g / 44.1 g/mol = 0.566 mol

Since 1 mole of propane reacts with 5 moles of oxygen, we can calculate the moles of oxygen:

Moles of oxygen = Moles of propane * 5 = 0.566 mol * 5 = 2.83 mol

Therefore, the air/fuel ratio is 2.83 mol of oxygen / 0.566 mol of propane = 5.