Question

If 88.0 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at 720. mm Hg and 22ºC, how many grams of H2O?

Answer 1

126 grams of H2O is formed.

Step-by-step explanation:

Data given:

volume of the gas = 88 Liters

pressure = 720 mm Hg or 0.947 atm

temperature T = 22 Degrees or 295.15 K

R = 0.08021 atm L/mole K

n =?

The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.

PV = nRT

n =
`(PV)/(RT)`

putting the values in the equation

= 0.947 X 88/ 0.08021 X 295.15

n = 3.5 moles

balanced reaction for combustion of methane

CH4 + O2 ⇒ CO2 + 2H20

1 mole of CH4 undergoes combustion to form 2 moles of water

3.5 moles will give x moles of water

2/1 = x/3.5

x = 7 moles of water (atomic mass of water = 18 gram/mole)

mass = atomic mass x number of moles

mass = 18 x 7

=126 grams of water is formed.