<h3>
Answer:</h3>
45.57 g of LiO
<h3>
Explanation:</h3>
We are given;
Mass of Lithium = 12.2 g
Mass of oxygen = 12.2 g
We are required to calculate the theoretical yield of the reactant.
The reaction between Lithium and oxygen is given by;
2Li(s) + O₂(g) → 2LiO(s)
<h3>Step 1: Determine the rate limiting reagent</h3>
- We need to determine the number of moles of Lithium and Oxygen to determine the rate limiting reagent.
Moles of Lithium;
Moles = Mass ÷ Molar mass
Molar mass of Lithium = 6.941 g/mol
Therefore;
Moles of Lithium = 12.2 g ÷ 6.941 g/mol
= 1.758 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of oxygen = 12.2 g ÷ 16.0 g/mol
= 0.7625 moles
From the reaction 2 mole of Li reacts with 1 mole of oxygen to produce 2 moles of LiO.
Therefore, Oxygen is the limiting reagent because it has fewer number of moles
<h3>Step 2: Calculating the number of moles </h3>
1 mole of oxygen reacts to produce 2 moles of LiO.
Therefore, the mole ratio of oxygen to Lithium oxide is 1 : 2
Thus, moles of LiO = Moles of oxygen × 2
= 0.7625 moles × 2
= 1.525 moles of LiO
<h3>Step 3: Calculating the theoretical mass of LiO</h3>
Mass = number of moles × Molar mass
Molar mass of LiO = 29.88 g/mol
Therefore;
Mass of LiO = 1.525 moles × 29.88 g/mol
= 45.567 g
= 45.57 g
Therefore, the theoretical mass of the product, LiO is 45.57 g
