Answer:
63.73 g KClO₃
Step-by-step explanation:
We are given the reactants;
- 115 g of KCl
- 25 g of oxygen
We are required to determine the theoretical yield;
First we write the equation for the reaction;
2KCl(s) + 3O2(g) → 2KClO3(s)
Second, we determine the number of moles of each reactant;
Moles of KCl = 115 g ÷ 74.55 g/mol
= 1.54 moles
Moles of Oxygen = 25 g ÷ 32.0 g/mol
= 0.78 moles
Thirdly, we determine the rate limiting reagent;
From the equation 2 moles of KCl reacts with 3 moles of O₂
Therefore, 1.54 moles of KCl will require 2.31 moles of O₂, which means KCl is in excess.
Therefore, Oxygen is the rate limiting reagent.
Fourth step: Determine the number of moles of the product KClO₃
From the equation, 3 moles of O₂ reacts to produce 2 moles of KClO₃
Therefore;
Moles of KClO₃ = Moles of O₂ × 2/3
= 0.78 moles × 2/3
= 0.52 moles of KClO₃
Fifth step; Determine the theoretical yield of reaction
Moles of KClO₃ produced = 0.52 moles
But; Mass = Number of moles × molar mass
Therefore;
Mass of KClO₃ = 0.52 moles × 122.55 g/mol
= 63.726 g
= 63.73 g
Therefore, the theoretical yield of the reaction is 63.73 g KClO₃