Question

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

• A. 132g
• B. 44g
• C. 88g
• D. 176g

Hint:

A quick shortcut: Since $16 \text{ g}$ of $CH_4$ (1 mole) gives $44 \text{ g}$ of $CO_2$ (1 mole), doubling the reactant to $32 \text{ g}$ (2 moles) will naturally double the product mass to $88 \text{ g}$ of $CO_2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question requires us to determine the mass of carbon dioxide ($CO_2$) gas formed when $32 \text{ g}$ of methane ($CH_4$) gas is completely burned in the presence of excess (sufficient) oxygen.

Step 2: Key Formula or Approach:
First, write the balanced chemical equation for the combustion of methane: \[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g) \] According to the stoichiometry of this reaction, $1 \text{ mole}$ of $CH_4$ reacts to produce exactly $1 \text{ mole}$ of $CO_2$.
The molar mass of methane ($CH_4$) is: \[ M(CH_4) = 12 + 4 \times 1 = 16 \text{ g/mol} \] The molar mass of carbon dioxide ($CO_2$) is: \[ M(CO_2) = 12 + 2 \times 16 = 44 \text{ g/mol} \]

Step 3: Detailed Explanation:

• Calculate the moles of $CH_4$ reactant starting with $32 \text{ g}$:
\[ \text{Moles of } CH_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ moles} \]

• From the balanced chemical equation, the mole ratio between $CH_4$ and $CO_2$ is $1:1$.
Therefore, $2 \text{ moles}$ of $CH_4$ will produce:
\[ \text{Moles of } CO_2 \text{ produced} = 2 \text{ moles} \]

• Now, convert the moles of $CO_2$ into mass:
\[ \text{Mass of } CO_2 = \text{Moles of } CO_2 \times M(CO_2) \] \[ \text{Mass of } CO_2 = 2 \text{ moles} \times 44 \text{ g/mol} = 88 \text{ g} \]

• Thus, the combustion yields exactly $88 \text{ g}$ of $CO_2$ gas.

Step 4: Final Answer:
The amount of $CO_2$ gas produced from the combustion of $32 \text{ g}$ of methane is $88 \text{ g}$.