Question

Calculate the amount of \( CO_2 \) gas produced, when \( 32 \)g of \( CH_4 \) is burned with sufficient amount of oxygen (Given atomic weight of \( C=12, O=16 \) and \( H=1 \))

• A. \( 132 \)g
• B. \( 44 \)g
• C. \( 88 \)g
• D. \( 176 \)g

Hint:

In stoichiometry, always "bridge" through moles. Convert the given mass to moles, use the coefficients from the balanced equation to find the moles of the desired product, and then convert those moles back to mass.

Answer 1

The Correct Option is C

Concept: Stoichiometry allows us to calculate the masses of reactants and products in a chemical reaction. The combustion of methane (\( CH_4 \)) in excess oxygen is a complete combustion reaction. To solve this, we must first write the balanced chemical equation and then use the mole concept to relate the mass of the reactant to the mass of the product.

Step 1: Write the balanced chemical equation for the combustion of methane.
Methane reacts with oxygen to produce carbon dioxide and water: \[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \] From the equation, \( 1 \) mole of \( CH_4 \) produces \( 1 \) mole of \( CO_2 \).

Step 2: Calculate the number of moles of \( CH_4 \)
Molar mass of \( CH_4 = 12 + (4 \times 1) = 16 \) g/mol. \[ \text{Moles of } CH_4 = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ moles} \]

Step 3: Determine the mass of \( CO_2 \) produced.
Since the molar ratio of \( CH_4 \) to \( CO_2 \) is \( 1:1 \), \( 2 \) moles of \( CH_4 \) will produce \( 2 \) moles of \( CO_2 \). Molar mass of \( CO_2 = 12 + (2 \times 16) = 44 \) g/mol. \[ \text{Mass of } CO_2 = \text{Moles} \times \text{Molar mass} = 2 \text{ moles} \times 44 \text{ g/mol} = 88 \text{ g} \]