To find the mass of aluminium oxide (Al2O3) produced, we start with the balanced chemical equation for the reaction:
4Al + 3O2 → 2Al2O3
1. **Molar Mass Calculation:**
- Al (Aluminium) = 27 g/mol
- O2 (Oxygen) = 32 g/mol
- Al2O3 (Aluminium oxide) = (2×27 + 3×16) g/mol = 102 g/mol
2. **Determine Limiting Reagent:**
The moles of Al from 81.0 g:
n(Al) = 81.0 g / 27 g/mol = 3.0 mol
The moles of O2 from 128.0 g:
n(O2) = 128.0 g / 32 g/mol = 4.0 mol
According to the equation, 4 moles of Al react with 3 moles of O2. Calculate the required oxygen for 3.0 mol of Al:
Required O2 = (3.0 mol Al) × (3/4) = 2.25 mol O2
Since 2.25 mol of O2 is less than the available 4.0 mol, Al is the limiting reagent.
3. **Calculate Mass of Al2O3 Produced:**
From 3.0 mol of Al (1.5 mol of Al2O3):
n(Al2O3) = 3.0 mol × (2/4) = 1.5 mol
Mass of Al2O3 = 1.5 mol × 102 g/mol = 153.0 g
The mass of aluminium oxide produced is 153 g.
This value falls within the expected range of 153 to 153 g.
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Match the LIST-I with LIST-II for an isothermal process of an ideal gas system. 
Choose the correct answer from the options given below:
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?
