The balance chemical equation is as follow,
2 Al + 3 O₂ → Al₂O₃
Aluminium is the Limiting Reagent,
As,
107.92 g Al required = 96 g of O₂
Then,
82.49 g of Al will require = X g of O₂
Solving for X,
X = (82.49 g × 96 g) ÷ 107.92 g
X = 73.37 g of O₂
But,
We are provided with 117.65 g of O₂, So, it is provided in excess and 44.28 g of it will remain unreacted.
Solving for Amount of Al₂O₃ formed,
As,
107.92 g of Al produced = 203.92 g of Al₂O₃
Then,
82.49 g of Al will produce = X g of Al₂O₃
SOlving for X,
X = (82.49 g × 203.92 g) ÷ 107.92
X = 155.86 g of Al₂O₃
Answer:
There are 0.301 moles of N2 gas in 6.75 liters of N2 gas.
Explanation:
Given,
Volume of N2 gas = 6.75 L
We know,
At STP,
Number of moles = 
[Here 22.4 liter is the molar volume of any gas at STP]
= 
= 
∴There are 0.301 moles of N2 gas in 6.75 liters of N2 gas.
Yes as it is less dense than the liquid water
5. Eubacteria
6. Plantae
7. Animalia
8. Protist (technically not a kingdom)
9. Archaebacteria
10. Fungi
