Answer :
Final answer:
The number of grams of oxygen required for the complete combustion of 4.00 grams of methane is 16.0 grams (option c). This is found by calculating the moles of methane and using the stoichiometric relationship to determine the moles and then the mass of oxygen required.
Explanation:
The question asks for the number of grams of oxygen required for the complete combustion of 4.00 grams of methane (CH4). The balanced chemical equation for the combustion of methane is CH4 + 2 O2 → CO2 + 2 H2O. To solve this question, we need to use stoichiometry based on the balanced equation.
First, we calculate the number of moles of CH4 in 4.00 grams:
- Molar mass of CH4 = 12.01 g/mol (carbon) + 4.00 g/mol (hydrogen) = 16.01 g/mol
- Moles of CH4 = mass/molar mass = 4.00 g / 16.01 g/mol = 0.25 moles
According to the balanced equation, 1 mole of CH4 requires 2 moles of O2. Thus, 0.25 moles of CH4 requires 0.50 moles of O2.
- Molar mass of O2 = 32.00 g/mol
- Mass of O2 required = moles × molar mass = 0.50 moles × 32.00 g/mol = 16.0 grams
Therefore, the number of grams of oxygen required for the combustion of 4.00 grams of methane is 16.0 grams (option C).
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