Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
Answer:
I don't really get the options but it favoures the reactant side.
Explanation:
Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g
If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.
With the explanations I have made, I hope the question is now clear to you.
Step 1 : Write balanced chemical equation.
CaF₂ can be converted to F₂ in 2 steps. The reactions are mentioned below.
I] 
II] 
The final balanced equation for this reaction can be written as
Step 2: Find moles of CaF₂ Using balanced equation
We have 1.12 mol F₂
The mole ratio of CaF₂ and F₂ is 1:1
Step 3 : Calculate molar mass of CaF2.
Molar mass of CaF₂ can be calculated by adding atomic masses of Ca and F
Molar mass of CaF₂ = Ca + 2 (F)
Molar mass of CaF₂ = 40.08 + 18.998 = 78.08 g
Step 4 : Find grams of CaF₂
Grams of CaF₂ = 
Grams of CaF₂ = 87.45 g
87.45 grams of CaF2 would be needed to produce 1.12 moles of F2.
