Question

What volume of air is needed to burn completely 100cm of propane?

C3H8(g) + 5O2 —-> 3CO2 + 4H2O

Answer 1

See below

Step-by-step explanation:

propane mole weight = 44 gm / mole

100 gm / 44 gm / mole = 2.27 moles

From the equation, 5 times as many moles of OXYGEN (O2)are required

= 11.36 moles of oxygen

at STP this is 254.55 liters of O2 (because 22.4 L = one mole) and

Using oxygen as 21 percent of air means that

.21 x = 254.55 = x = 1212.12 liters of air required