Answer:
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
Answer: The volume of oxygen gas needed is 156.8 L
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of methane = 56 g
Molar mass of methane = 16 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of methane}=\frac{56g}{16g/mol}=3.5mol[/tex]
The chemical equation for the combustion of methane follows:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
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 = [tex]\frac{2}{1}\times 3.5=7mol[/tex] of oxygen gas
At STP:
1 mole of a gas occupies 22.4 L of volume
So, 7 moles of oxygen gas occupies [tex]\frac{22.4}{1}\times 7=156.8L[/tex] of volume
Hence, the volume of oxygen gas needed is 156.8 L
