If 12.2 g of lithium reacts with 12.2 g of oxygen what is the theoretical yield of the reaction in grams

45.57 g  of LiO

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)

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

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

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

idk what the question is about