Question

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

Answer 1

Answer: The volume of oxygen gas needed is 156.8 L

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of methane = 56 g

Molar mass of methane = 16 g/mol

Putting values in above equation, we get:

`\text{Moles of methane}=(56g)/(16g/mol)=3.5mol`

The chemical equation for the combustion of methane follows:

`CH_4+2O_2\rightarrow CO_2+2H_2O`

By Stoichiometry of the reaction:

1 mole of methane reacts with 2 moles of oxygen gas

So, 3.5 moles of methane will react with =
`(2)/(1)* 3.5=7mol` of oxygen gas

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 7 moles of oxygen gas occupies
`(22.4)/(1)* 7=156.8L` of volume

Hence, the volume of oxygen gas needed is 156.8 L

Answer 2

The volume of oxygen will be 168 dm³.

Step-by-step 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