How many molecules of water will be produced if 13.15 g of methane are burned?

Burning of methane is a combustion reaction. CH₄(g) reacts with O₂(g) and gives CO₂(g) and H₂O(l) as the products.

The balance reaction equation is 
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) 

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of CH₄ = 13.15 g 
Molar mass of CH₄ = 16.04 g/mol
Moles of CH₄ =  13.15 g /16.04 g/mol
                      = 0.82 mol
                        

The stoichiometric ratio between CH₄ and H₂O is 1 : 2.
Hence,
          moles of H₂O formed = moles of reacted CH₄ x 2
                                               = 0.82 x 2
                                               = 1.64 mol

According to the Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.

Hence,
    number of H₂O molecules formed = 1.64 mol x 6.022 × 10²³ mol

                                                              = 9.88 x 10²³

Hence, answer is 9.88 x 10²³ of molecules.

Less stable  hope this

Where is the following?