Question

How many liters of oxygen gas can react with 84.0 grams of lithium metal at standard temperature and pressure? Show all of the work used to find your answer. 4Li + O2 yields 2Li2O

Answer 1

Answer: The volume of oxygen gas reacted is 67.2 L.

Explanation:

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

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

For Lithium

Given mass of lithium = 84 g

Molar mass of lithium = 7 g/mol

Putting values in above equation, we get:

$\text{Moles of lithium}=\frac{84g}{7g/mol}=12mol$

For the given chemical reaction:

$4Li+O_2\rightarrow 2Li_2O$

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 = $\frac{1}{4}\times 12=3moles$ of oxygen gas.

At STP:

1 mole of a gas occupies 22.4 L of gas.

So, 3 moles of a gas will occupy = $3\times 22.4=67.2L$

Hence, the volume of oxygen gas reacted is 67.2 L.

Answer 2

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.