Respuesta :
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: [tex]n = \frac{given\: mass (m)}{molar\: mass (M)}[/tex]
∴ Number of moles of CH₄: [tex]n_{1} = \frac{m_{1}}{M_{1}} = \frac{1 g}{16 g} = 0.0625 mol[/tex]
Number of moles of O₂: [tex]n_{2} =\frac {m_{2}}{M_{2}} = \frac{3 g}{32 g} = 0.0937 mol[/tex]
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 [tex]\frac{2 \times 0.0625}{1} = 0.125 mol \: O_{2}[/tex].
Thus the limiting reagent is O₂.
Now, 2 moles O₂ gives 1 mole of CO₂
So 0.0937 mol O₂ gives [tex]\frac{1 \times 0.0937}{2} = 0.0468 mol \:CO_{2}[/tex]
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: [tex]PV =nRT[/tex]
[tex]\Rightarrow V = \frac{nRT}{P}[/tex]
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:
[tex]V = \frac{n_{3}RT}{P}[/tex]
[tex]V = \frac{0.0468 mol \times 0.08206 L.atm/(mol.K) \times 298K}{ 1atm}[/tex]
[tex]\Rightarrow V = 1.14 L[/tex]
Therefore, the total volume of the balloon after the completion of the reaction: V = 1.14 L
