Question

A propane gas (C3H8) heater burns propane at a rate of 500 grams per hour.
a) What is the oxygen consumption rate? (in g O2/hr.)
b) If the heater uses a blower to provide outside air (0°C, 1 atm) for combustion, what is the minimum air flow rate the blower must provide? (in L/min)

Answer 1

start with the balanced chem equation
C3H8 + 5O2 >>3CO2 + 4H2O

so the ratio of moles is 5O2/1C3H8
and the ratio of mass is 5*32gO2/44g propane
so for every 500 g propane/perhour, one used 500*(44g) O2/hour
Outside air is only 27 percent Oxygen, one needs air flow rate
airflowrate=500*44gO2*22.4liters/32gO2 * 1/.27= ...