Question

The following is the balanced equation of the combustion of methane gas with oxygen gas:

Answer 1

Answer:

The correct option is: C. 1.1 L

Explanation:

Given: Temperature: T = 25°C = 25 +273 = 298K      (∵ 1°C = 273K)

Pressure: P = 1 atm, Given mass of- CH₄: m₁ = 1g; O₂: m₂ = 3g

Volume: V= ?

Molar mass of CH₄: M₁ = 16g; O₂: M₂ = 32g

As Number of moles: $n = \frac{given\: mass (m)}{molar\: mass (M)}$

∴ Number of moles of CH₄: $n_{1} = \frac{m_{1}}{M_{1}} = \frac{1 g}{16 g} = 0.0625 mol$

Number of moles of O₂: $n_{2} =\frac {m_{2}}{M_{2}} = \frac{3 g}{32 g} = 0.0937 mol$

In the given reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

1 mole of methane (CH₄) reacts with 2 moles oxygen (O₂) to give 1 mole of CO₂.

So 0.0625 mol of CH₄ reacts with $\frac{2 \times 0.0625}{1} = 0.125 mol \: O_{2}$.

Thus the limiting reagent is O₂.

Now, 2 moles O₂ gives 1 mole of CO₂

So 0.0937 mol O₂ gives $\frac{1 \times 0.0937}{2} = 0.0468 mol \:CO_{2}$

Therefore, the total number of moles of gas after the completion of the reaction: n = number of moles of CO₂: n₃ = 0.0468 mol

Now to calculate the volume of the balloon, we use the ideal gas law: $PV =nRT$

$\Rightarrow V = \frac{nRT}{P}$

Here, R is the gas constant = 0.08206 L·atm/(mol·K)

T is Temperature,

P is Pressure,

n is Total number of moles of gas

and, V is Volume

Therefore, the volume of the balloon after the completion of the reaction:

$V = \frac{n_{3}RT}{P}$

$V = \frac{0.0468 mol \times 0.08206 L.atm/(mol.K) \times 298K}{ 1atm}$

$\Rightarrow V = 1.14 L$

Therefore, the total volume of the balloon after the completion of the reaction: V = 1.14 L