Question

How many liters of hydrogen gas can be produced if 102 grams of methane gas (CH) are combusted at 315 K and 1.2 atm? Show all of the work used to solve this problem.
CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)

Answer 1

find moles, using molar mass
(102 g CH4) ( 1mol / 16.04 g CH4) = 6.359 mol CH4

by the equation:
CH4 (g) + 2 O2 (g) --> CO2 (g) + 2 H2O (g)
6.359 mol CH4 produces twice as many moles of H2O = 12.72 moles H2O

PV = nRT
(1.2 atm) V = (12.72 moles H2O) (0.08206 L-atm/mol-K) (315 K)
V = 274 litres of water vapor

Answer 2

Thanks. (:

Answer 3

np(: