Question

How many molecules of \(CO_2(g)\) are obtained on reaction of 24 grams of methane with 4 moles of oxygen?

• A. \(6.022 \times 10^{23}\)
• B. \(3.011 \times 10^{23}\)
• C. \(12.044 \times 10^{23}\)
• D. \(9.033 \times 10^{23}\)

Hint:

Always identify the limiting reagent first in stoichiometry problems. The limiting reagent determines the amount of product formed.

Answer 1

The Correct Option is D

Concept: This is a stoichiometry problem involving combustion reaction and limiting reagent concept.

Step 1: Writing the balanced chemical equation.
Combustion of methane occurs as: \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] From the balanced equation: \[ 1 \text{ mole of } CH_4 \rightarrow 1 \text{ mole of } CO_2 \]

Step 2: Calculating moles of methane.
Molar mass of methane: \[ CH_4 = 12 + 4(1) = 16 \text{ g/mol} \] Given mass: \[ 24 \text{ g} \] Moles of methane: \[ \frac{24}{16} = 1.5 \text{ moles} \]

Step 3: Checking limiting reagent.
According to equation: \[ 1 \text{ mole } CH_4 \text{ requires } 2 \text{ moles } O_2 \] Thus: \[ 1.5 \text{ moles } CH_4 \text{ require } 3 \text{ moles } O_2 \] Available oxygen: \[ 4 \text{ moles} \] Since oxygen available is greater than required: \[ CH_4 \] is the limiting reagent.

Step 4: Calculating moles of carbon dioxide formed.
From stoichiometry: \[ 1 \text{ mole } CH_4 \rightarrow 1 \text{ mole } CO_2 \] Therefore: \[ 1.5 \text{ moles } CH_4 \rightarrow 1.5 \text{ moles } CO_2 \]

Step 5: Calculating number of molecules.
Using Avogadro number: \[ N_A = 6.022 \times 10^{23} \] Number of molecules: \[ 1.5 \times 6.022 \times 10^{23} \] \[ = 9.033 \times 10^{23} \] Hence, the number of molecules formed is: \[ 9.033 \times 10^{23} \]