Calculating maximum mass

Question

Consider the reaction CH4 +  2 O2    CO2 +  2 H2O.

What is the maximum mass  (in grams) of H2O that can be produced if 5.92 g of CH4 reacts with excess O2?

Answer 1

Hi, Noreen,

For this problem, let's be sure that the equation provided is indeed balanced:

CH₄ + 2O₂ -> CO₂ + 2 H₂O

If we check, we'll find one carbon on each side, four H's on each side, and four oxygen atoms on each side. So, that's good, it is a balanced reaction!

Next, to answer the question (and since it is stated that oxygen is in excess), we need to first convert 5.92 grams of methane (CH₄) into moles. This is done by recognizing the following formula:

X moles CH₄ = Weight of CH₄

Molecular weight of methane

= 5.92 gm CH₄

16.04 g/mol

Therefore, 5.92 grams of methane is equal to 0.37 moles of methane.

Solving the equation above:

CH₄ + 2O₂ -> CO₂ + 2 H₂O, so

0.37 moles CH₄ + excess O₂ -> 0.37 moles CO₂ + 0.74 moles H₂O

Finally, 0.74 moles of water is equal to (0.74 moles)(18.02 g/mol) or 13.3 grams of H₂0 is the maximum mass of water produced under these conditions.

'Hope that helps,

Chris