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
<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams
<u>Explanation:</u>
We are given:
44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.
To calculate volume of a substance, we use the equation:
Density of solution = 1.343 g/mL
Mass of solution = 100 g
Putting values in above equation, we get:
To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:
In 77.46 mL of solution, mass of sulfuric acid present is 44 g
So, in 60 mL of solution, mass of sulfuric acid present will be = 
Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams
Answer:
c
O2: 7 mol
CO2: 4 mil
