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?

126 grams of H2O is formed.

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 =

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.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-