Question

When propane is burned, the balanced reaction is: C3H8(l) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g). If 31.66 L of water vapor are produced at 865.7 torr and 75°C, how many grams of propane were burned?

Answer 1

The reaction:

C3H8(l) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g).

is already balanced

so

relate mol of water to mol of propane

mol of water = from ideal gas law

Assume Pressure is pressure of GAS and not of device

so

Pgas = 865.7 torr

R = 62.4 tottL/molK

T = 75°C = 75+273 K = 348 K

V = 31.66 L

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

n = PV/(RT)

n = (865.7*31.66)/(62.4*348)

n = 1.262

1 mol of propane = 4 mol of water

x mol of propane = 1.262 mol of water

x = 1.262/4 = 0.3155 mol of propane

mass= mol*MW = 0.3155*44.10 = 13.91355 g of propane must be burnt