Question

if a .048 sample of pentane burns completely in O2 what will be the pressure of the water vapor in a 5.45 L flask at 28.5 C?

Answer 1

pentane formula = C5H12

C5H12 + 8 O2 ---------------------> 5 CO2 + 6 H2O

1 mole C5H12 = 6 mole of H2O

0.048 mole C5H12 = 6 x 0.048 / 1

                              = 0.288

moles of water = 0.288

volume = 5.45 L

temperature = 28.5 + 273 = 301.5 K

P V = n R T

P   x    5.45 = 0.288 x 0.0821 x 301.5

P = 1.31 atm

pressure of water vapour = 1.31 atm