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

ch4(g) + 2o2(g) -> co2(g) + 2h2o(l)
if you ran a reaction in a sealed balloon at 25o c with 1 g of methane, ch4, and 3 g of oxygen, o2, what is the end volume of the balloon once the reaction is complete? the pressure was constant at 1 atm. assume all gases behave ideally and you have 100% yield. (ignore vapor pressure of water)

a. 4.9 l
b. 0.096 l
c. 1.1 l
d 1.5 l
e. 0.13 l

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:

∴ Number of moles of CH₄:

Number of moles of O₂:

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 .

Thus the limiting reagent is O₂.

Now, 2 moles O₂ gives 1 mole of CO₂

So 0.0937 mol O₂ gives

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:

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:

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

1 is the answer to this problem