Question

a sample of methane gas having a volume of 2.8 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35L at 31°C and 1.25 atm. The mixture was ignited to form carbon dioxide and water. Calculate the volume of carbon dioxide formed at a pressure of 2.5 atm and a temperature of 125°C

Answer 1

→ Answer:

2.48 L

→ Explanation:

Equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Table of given conditions:

`\left[\begin{array}{cccc}&Volume(L)&Pressure(atm)&Temperature(K)\\Methane&2.8&1.65&298\\Oxygen&35&1.25&304\end{array}\right]`

note: K = C° + 273 (Temperature conversion, Celsius to Kelvin)

Find moles of methane and oxygen:

note:
`Ideal \ Gas \ Law: \ PV=nRT \ \ \ (R = 0.082 \ (L * atm )/(K * mol) )`

Formula to find moles is:
`n=(PV)/(RT)`

`\space \\CH_(4) \ : \ n = (1.65 * 2.8)/(298 * 0.082) = 0.19\ mol \ (limiting \ reactant)`

`O_(2) \ : \ n = (1.25 * 35)/(304 * 0.082) = 1.75\ mol \ (2 * 0.19 = 0.38 \ mol \ is \ enough)`

note: moles of CO₂ is the same as of CH₄, as can be seen from the chemical equation above, which is 0.19 mol.

Find volume of CO₂:

`\left[\begin{array}{ccccc}&Volume(L)&Pressure(atm)&Temperature(K)&Moles(mol)\\CO_(2) &?&2.5&398&0.19\end{array}\right]`

Formula to find Volume is:
`V=(nRT)/(P)`

`CO_(2) \ : \ V = (0.19 * 0.082 * 398)/(2.5) = 2.48 \ L`

So, the answer is: 2.48 L