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.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 