Question

Methane burns in the presence of oxygen to produce carbon dioxide and water in the following reaction:

CH4 (s) + 2 O2 (g) --> CO2 (g) + 2 H2O (l)

What mass of methane (in grams) will require 35.15 L of oxygen to fully react? The pressure is 2.35 atm and the temperature is 15 °C. (R = 0.0821 L atm/mol K)

Answer 1

Answer:

83.76g of Methane

Explanation:

From the equation of reaction

CH4 (s) + 2 O2 (g) --> CO2 (g) + 2 H2O (l)

1 mole of methane is required to react with 2 moles of oxygen.

Volume of oxygen = 35.15L

Pressure = 2.35 atm

T = 15°C = (15 + 273.15)k = 288.15K

Number of moles of oxygen = 2

R = 0.0821 L atm/mol K

From ideal gas equation,

PV = nRT

n = P*V / RT

n = (2.35 * 35.15) / (0.0821 * 288.15)

n = 82.60 / 23.657

n = 3.49 moles.

From the stoichiometry of the reaction,

1 mole of CH4 = 2 moles of O2

X moles of CH4 = 3.49 moles of O2

X = (3.49 * 1) / 2

X = 1.745 moles

Now, we can apply mole - mass relationship

Number of moles = mass / molarmass

Molar mass of CH4 = [12 + (1*4)] = 48 g/mol

Mass = number of moles * molarmass

Mass = 1.745 * 48

Mass = 83.76g

The mass of methane required to react with that amount of oxygen is 83.76g

Answer 2

Answer:

The mass of methane that will require 35.15 L of oxygen to fully react is 27.04 grams

Explanation:

Here we have 2 moles of oxygen gas combining with 1 mole of methane gas

At 2.35 atm  and 15 °C, the volume of oxygen is 35.15 L

Number of moles, n of oxygen is given by the following relation;

$n = \frac{PV}{RT}$

Therefore

$n = \frac{2.35 \times 35.15}{0.0821 \times 298.15} = 3.375 \ moles$

Therefore, since 2 moles of oxygen gas combining with 1 mole of methane the number of moles of methane combining with 3.375 moles of oxygen = 3.375/2 moles or 1.69 moles of methane

Molar mass of methane = 16.04 g/mol

Therefore, mass of 1.69 moles of methane = 16.04 g/mol × 1.69 mole = 27.064 g

Which means that the mass of methane that will require 35.15 L of oxygen to fully react = 27.04 grams.