Question

7. What volume of oxygen is required to react with 14.0 liters of propane gas to produce carbon dioxide and water? (unbalanced) C 3 H 8 + O 2 → CO 2 + H 2 O

8. The volume of a sample of gas measured at 26 C and 0.986 atm is 10.50 liters. What final temperature, in degrees Celsius, would be required to reduce the volume to 9.50 liters at constant pressure?

Answer 1

7. Balanced equation

C3H8 + 5O2 ------> 3CO2 + 4H2O

We know, volume is directly proportional to number of moles.

From reaction, 1.0 mol of C3H8 reacts with 5.0 mol of O2 so volume of oxygen gas required = 5.0 * 14.0 Liters = 70.0 liters

8. At constant pressure, V1 / T1 = V2/T2 ----(1)

V1 = 10.50 Liters, T1 = 26 oC = (26 + 273) K = 299 K

V2 = 9.50 Liters, T2 = ?

From (1)

T2 = V2 * T1 / V1 = 9.50 Liters * 299 K / 10.50 Liters =270.52K

T2 = (270.52 - 273) oC = -2.48 oC

Final temperature = T2 = - 2.48 oC