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

when the balanced reaction equation for the reaction of methane gas and O2 is:

CH4(g) + 2O2(g)  → CO2(g) + 2H2O(g)

we will use the ideal gas equation to get the moles of each gas:

1) moles of methane:

PV= n RT 

when p is the pressure =1.65 atm

V is the volume = 2.8 L

R is the ideal gas constant = 0.0821

T is the temperature in Kelvin = 25+ 273=  298 K

∴n of CH4 = 1.65 * 2.8 / 0.0821 * 298 K = 0.189 mole

2) moles of O2:

PV = nRT

when P = 1.25 atm

and V= 35L

and R = 0.0821 L atm / mol K

T = 31 + 273 = 304 K

so, by substitution:

n of O2= 1.25 atm * 35 L / 0.0821 * 304 = 1.75 mol

from the balanced reaction equation, we can see that the molar ratio between O2 and CH4 is 2: 1 

∴ moles of O2 = 2 * .0189 moles of methane = 0.378 mole

so the O2 exists in excess and CH4 is the limiting reactant

when 1 mole of CH4 → 1 mole ofCO2

 1 mol of CO2 → 22.4 L CO2

 volume of CO2 at STP =( 0.189 molesCH4 /1)*(1mol CO2)/(1 mole CH4)*(22.4L CO2)/(1molCO2) = 4.23 L

then we will use this formula to get the correct value of V of CO2 at laboratory conditions:

P1V1 /T1 = P2V2/T2


when at STP conditions:

P1= 1 atm 

V1 = 4.23 L 

T1 = 273

and the laboratories conditions are:

P2 = 2.5 atm 

T2 = 125 + 273 = 398 K

∴ V2 = 1 atm * 4.23 L *398 / 273 * 2.5 atm

∴ V2 = 15.4 L

∴ the volume of CO2 formed at pressure of 2.5 atm and T = 125°C is 15.4 L

Answer 2

→ 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:  $Ideal \ Gas \ Law: \ PV=nRT \ \ \ (R = 0.082 \ \frac{L \times atm }{K \times mol} )$

Formula to find moles is:  $n=\frac{PV}{RT}$

$\space \\CH_{4} \ : \ n = \frac{1.65 \times 2.8}{298 \times 0.082} = 0.19\ mol \ (limiting \ reactant)$

$O_{2} \ : \ n = \frac{1.25 \times 35}{304 \times 0.082} = 1.75\ mol \ (2 \times 0.19 = 0.38 \ mol \ is \ enough)$

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:  $V=\frac{nRT}{P}$

$CO_{2} \ : \ V = \frac{0.19 \times 0.082 \times 398}{2.5} = 2.48 \ L$

So, the answer is:   2.48 L