Step 1: Understanding the Concept:
This problem is a classic application of Stoichiometry in chemical reactions.
Stoichiometry allows us to predict the amount of product formed in a reaction based on the amount of reactants provided.
The combustion of methane (\(CH_4\)) is a highly exothermic reaction where the hydrocarbon reacts with oxygen (\(O_2\)) to produce carbon dioxide (\(CO_2\)) and water vapor (\(H_2O\)).
The phrase "sufficient amount of oxygen" tells us that oxygen is in excess.
Therefore, methane is the "limiting reactant," and the amount of product formed depends entirely on how much methane is available to react.
To solve this, we must follow a three-step path: Mass of reactant \(\rightarrow\) Moles of reactant \(\rightarrow\) Moles of product \(\rightarrow\) Mass of product.
Key Formula or Approach:
1. Write and balance the chemical equation:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]
This balanced equation shows a 1:1 molar ratio between \(CH_4\) and \(CO_2\).
2. Determine Molar Masses:
Molar mass of \(CH_4 = 12 + (4 \times 1) = 16 \text{ g/mol}\).
Molar mass of \(CO_2 = 12 + (2 \times 16) = 44 \text{ g/mol}\).
Step 2: Detailed Explanation:
Let's perform the step-by-step calculation:
1. Convert the given mass of methane into moles:
Given mass of \(CH_4 = 32 \text{ g}\).
Number of moles of \(CH_4 = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ moles}\).
2. Use the stoichiometric ratio from the balanced equation:
From the reaction: 1 mole of \(CH_4\) produces 1 mole of \(CO_2\).
Since we have 2 moles of \(CH_4\), the reaction will produce exactly 2 moles of \(CO_2\).
3. Convert the moles of carbon dioxide into mass:
Mass of \(CO_2 = \text{Moles of } CO_2 \times \text{Molar Mass of } CO_2\)
Mass of \(CO_2 = 2 \text{ moles} \times 44 \text{ g/mol} = 88 \text{ g}\).
This calculation demonstrates the Conservation of Mass within the context of chemical identities. While we started with 32g of methane, the addition of oxygen atoms from the atmosphere (which weren't counted in the initial 32g) results in a much heavier product of 88g. The stoichiometry ensures that the number of carbon atoms remains constant (1 atom in methane becomes 1 atom in carbon dioxide).
Step 3: Final Answer:
The combustion of 32g of methane produces 88g of carbon dioxide.
This corresponds to option (C).