Asample of methane gas having a volume of 2.80 l at 25 degree c and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 l at 31 degree c and 1.25 atm. the mixture was then ignited to form carbon dioxide and water. calculate the volume of co_2 formed at a pressure of 2.50 atm and a temperature of 125 degree c.

Volume of carbon dioxide = 2.47 L

Explanation:

Given that:

For methane gas:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25.0 + 273.15) K = 298.15 K

V = 2.80 L

Pressure = 1.65 atm

Using ideal gas equation as:

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

1.65 atm × 2.80 L = n ×0.0821 L atm/ K mol  × 298.15 K  

⇒n for methane gas = 0.1887 mol

For oxygen gas:-

Temperature = 31 °C

T = (31 + 273.15) K = 304.15 K

V = 35.0 L

Pressure = 1.25 atm

Using ideal gas equation as:

Applying the equation as:

1.25 atm × 35.0 L = n ×0.0821 L atm/ K mol  × 304.15 K  

⇒n for oxygen gas = 1.7521 mol

According to the reaction:-

1 mole of methane gas reacts with 2 moles of oxygen gas

Also,

0.1887 mole of methane gas reacts with 2*0.1887 moles of oxygen gas

Moles of oxygen gas = 0.3774 moles

Available moles of oxygen gas  = 1.7521 moles

Limiting reagent is the one which is present in small amount. Thus, methane gas is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction forms 1 mole of carbon dioxide

Thus,

0.1887 mole of methane gas on reaction forms 0.1887 mole of carbon dioxide

Moles of carbon dioxide = 0.1887 moles

Given that:- Pressure = 2.50 atm

Temperature = 125 °C

T = (125 + 273.15) K = 398.15 K

Using ideal gas equation as:

Applying the equation as:

2.50 atm × V = 0.1887 moles ×0.0821 L atm/ K mol  × 398.15 K  

Volume of carbon dioxide = 2.47 L

law of conservation of matter