The balanced chemical equation for the combustion of methane (CH4) in the presence of oxygen (O2) is:
CH4 + 2O2 → CO2 + 2H2O
This equation tells us that for every one molecule of methane and two molecules of oxygen, we get one molecule of carbon dioxide and two molecules of water.
To find how many liters of water are produced by burning 38.25 grams of methane (CH4), we need to use stoichiometry. We can follow these steps:
Convert the given mass of CH4 to moles, using the molar mass of CH4:
38.25 g CH4 × (1 mol CH4/16.04 g CH4) = 2.38 mol CH4
Use the balanced chemical equation to find the number of moles of water produced:
1 mol CH4 produces 2 mol H2O
2.38 mol CH4 produces (2 mol H2O/mol CH4) × (2.38 mol CH4) = 4.76 mol H2O
Convert the number of moles of water to volume in liters, using the molar volume of a gas at standard conditions (STP):
1 mol of any gas at STP occupies 22.4 L
4.76 mol H2O occupies (22.4 L/mol) × (4.76 mol H2O) = 106.6 L H2O
Therefore, burning 38.25 grams of CH4 produces 106.6 liters of water at