If 1.320 moles of CH4 reacts completely with oxygen, how many grams of H2O can be formed

47.5 g of water can be formed

Explanation:

This is the reaction:

CH₄ + 2O₂ → CO₂ + 2H₂O

Methane combustion.

In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.

As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.

1.320 mol. 2 = 2.64 moles

Now, we can convert the moles to mass (mol. molar mass)

2.64 mol. 18g/mol = 47.5 g

mass H2O = 47.56 g

Explanation:

balanced reation:

CH4 + 2O2 → CO2 + 2H2O

∴ moles CH4 = 1.320 mol

⇒ moles H2O = (1.320 mol CH4) * (2 mol H2O/mol CH4)

⇒ moles H2O = 2.64 mol H2O

∴ molar mass (mm) H2O = 18.015 g/mol

⇒ mass H2O = (2.64 mol H2O) * (18.015 g/mol)

⇒ mass H2O = 47.56 g