if 20 grams of methane (ch4) reacts with 13 grams of oxygen, how many grams of water can be formed in this combustion reaction?

Answer 1

First, always write a correctly balanced equation:

CH₄ + 2O₂ ==> CO₂ + 2H₂O

Next, determine which reactant, if any, is limiting:

moles CH₄ = 20 g CH₄ x 1 mole/16 g = 1.25 moles CH₄

moles O₂ = 13 g O₂ x 1 mole/32 g = 0.41 moles O₂

Since it takes 2 moles O₂ for each mole of CH₄, clearly the O₂ is limiting, and it will determine how much H₂O can be formed.

Another way to determine which is limiting is to find moles of H₂O formed from each.

1.25 moles CH₄ x 2 moles H₂O/mole CH₄ = 1.25 moles H₂O formed

0.41 moles O₂ x 2 moles H₂O/2 moles O₂ = 0.41 moles H₂O formed = LIMITING

From the above calculation, you can make only 0.41 moles of H₂O, thus...

mass of H₂O formed = 0.41 moles H₂O x 18 g/mole = 7.3 g H₂O