1. The Chemical Equation
The reaction of methane (CH4) and oxygen (O2) is a combustion reaction, producing carbon dioxide (CO2) and water (H2O):
CH4 + 2O2 → CO2 + 2H2O
2. Stoichiometry
The equation tells us the following:
* 1 mole of CH4 reacts with 2 moles of O2
* 1 mole of CH4 produces 1 mole of CO2
* 1 mole of CH4 produces 2 moles of H2O
3. Molar Masses
* CH4: 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol
* O2: 2 * 16.00 g/mol (O) = 32.00 g/mol
4. Calculations
* Moles of CH4: 24 g / 16.05 g/mol = 1.5 mol
* Moles of O2: 96 g / 32.00 g/mol = 3 mol
5. Limiting Reactant
* From the equation: 1 mol of CH4 needs 2 mol of O2.
* We have: 1.5 mol of CH4 and 3 mol of O2.
* This means: We have enough O2 to react with all of the CH4. CH4 is the limiting reactant.
6. Product Calculations
* CO2 produced: 1.5 mol CH4 * (1 mol CO2 / 1 mol CH4) = 1.5 mol CO2
* Mass of CO2: 1.5 mol * 44.01 g/mol = 66.02 g
* H2O produced: 1.5 mol CH4 * (2 mol H2O / 1 mol CH4) = 3 mol H2O
* Mass of H2O: 3 mol * 18.02 g/mol = 54.06 g
Conclusion
When 24 g of CH4 reacts with 96 g of O2:
* CH4 is the limiting reactant.
* 66.02 g of CO2 and 54.06 g of H2O are produced.
