Question

Which starting condition would produce the larger volume of carbon dioxide by combustion of CH4(g) with an excess of oxygen gas to produce carbon dioxide and water: (a) 2.00 L of CH4(g); (b) 2.00 g of CH4(g)? Justify your answer. The system is maintained at a temperature of 75 C and 1.00 atm.

Answer 1

Answer:

(b) 2.00 g of CH4(g)

Explanation:

(a) 2.00 L of $CH_4$

Volume = 2.00 L

Temperature = 75°C

Pressure = 1.00 atm

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

T(K) = T( °C) + 273.15  

So,  

T = (75 + 273.15) K = 348.15 K  

T = 348.15 K

Using ideal gas equation as:

$PV=nRT$

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:

1.00 atm × 2.00 L = n × 0.0821 L.atm/K.mol ×  348.15 K

⇒n = 0.07 moles

(b) 2.00 g of $CH_4$

Mass of $CH_4$ = 2.00 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.00\ g}{16.04\ g/mol}$

Moles of $CH_4$ = 0.125 moles

Thus, more moles, more volume of carbon dioxide will be produced when more moles of methane react.

So, Answer - (b) 2.00 g of CH4(g)