Calculate the minimum volume of oxygen gas, at room temperature and pressure (r. t.p) required to completely burn 56g of methane gas

The volume of oxygen will be 168 dm³.

Explanation:

Given data:

mass of methane = 56 g

volume of oxygen = ?

Solution:

First of all we will write the balance chemical equation.

CH₄ + 2O₂ → CO₂ + 2H₂O

Now we will calculate the moles of methane:

number of moles = mass / molar mass

number of moles = 56 g / 16 g/mol

number of moles = 3.5 mol

Now we will compare the moles of oxygen and methane.

                                CH₄  :   O₂

                                   1     :     2  

                                 3.5   :    3.5 × 2 = 7

The moles of oxygen are 7 mol.

Now we will calculate the volume of oxygen. The gas is at room temperature and pressure so, one mole of gas will occupy 24 dm³ volume and 7 mole will occupy,

7 × 24 dm³ = 168 dm³

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"