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

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied