Answer :
From the equation:
4mol Li react with 1 mol O2
Molar mass Li = 7g/mol
mol in 84g Li = 84/7 = 12 mol Li
From the equation - 12 mol Li will react with 3 mol O2
At STP 1 mol O2 has volume = 22.4L
At STP 3 mol O2 has volume = 3*22.4 = 67.2L O2 gas will react.
4mol Li react with 1 mol O2
Molar mass Li = 7g/mol
mol in 84g Li = 84/7 = 12 mol Li
From the equation - 12 mol Li will react with 3 mol O2
At STP 1 mol O2 has volume = 22.4L
At STP 3 mol O2 has volume = 3*22.4 = 67.2L O2 gas will react.
Answer: The volume of oxygen gas reacted is 67.2 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]
For Lithium
Given mass of lithium = 84 g
Molar mass of lithium = 7 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of lithium}=\frac{84g}{7g/mol}=12mol[/tex]
For the given chemical reaction:
[tex]4Li+O_2\rightarrow 2Li_2O[/tex]
By Stoichiometry of the reaction:
4 moles of lithium reacts with 1 mole of oxygen gas.
So, 12 moles of lithium metal will react with = [tex]\frac{1}{4}\times 12=3moles[/tex] of oxygen gas.
At STP:
1 mole of a gas occupies 22.4 L of gas.
So, 3 moles of a gas will occupy = [tex]3\times 22.4=67.2L[/tex]
Hence, the volume of oxygen gas reacted is 67.2 L.