Question

Q1. For the reaction
Al(s) +O2 (g) => Al2O3 (s) (2)
i. Calculate the mass of aluminum oxide formed when 20 g of Al is burnt.
ii. How much oxygen gas is required to react with 40 g of Al?

Answer 1

Answer:

first calculate mass of aluminum oxide then minus it by 20

Answer 2

Given reaction,

⇒ Al + O₂ ➞ Al₂O₃

Balancing the reaction, we get

⇒ 4Al + 3O₂ ➞ 2Al₂O₃

[i]

We have to find the mass of aluminum oxide formed when 20 g or Al is burnt.

First, from the given reaction, we get to know that

⇒ 4 mole of Al produces 2 mole of Al₂O₃

So,

⇒ 1 mole of Al will produce 1/2 mole of Al₂O₃

So, Let us find the number of moles of Al,

⇒ No. of mole = Mass given / Molar mass

⇒ No. of mole = 20 / 26

⇒ No. of mole = 0.74 mole

But, One mole of Al produces 1/2 mole of Al₂O₃,

∴ 0.76 would produce = 0.76 × 1/2 mole

⇒ mole of Al₂O₃ = 0.38 mole

Now, We know

⇒ Moler mass of Al₂O₃ = 26 × 2 + 16 × 3

⇒ Molar mass = 52 + 48

⇒ Molar mass = 100 g

Finally, Mass of Aluminum oxide,

⇒ No. of moles = Mass / molar mass

⇒ 0.38 = mass / 100

⇒ mass = 38 g

Hence, Mass of Aluminum oxide produced is 38 g.

[ii]

Here, We need to find the mass of Oxygen gas required to react with 40 g of Al,

From the following reaction, It is known that

⇒ 3 mole of Oxygen gas would react with 4 mole of Al

Hence, 1 mole of Oxygen gas would react with 3/4 mole of oxygen gas.

Let us find mole of Aluminum,

⇒ No. of mole = Mass / Molar mass

⇒ No. of mole = 40 / 26

⇒ No. of mole of Al = 1.53 mole

So, Mole of oxygen gas required,

⇒ 1.53 × 3/4

⇒ 1.53 × 0.75

⇒ 1.14 mole

Now, Let us calculate the mass of Oxygen gas

⇒ No. of mole = Mass / Molar mass

⇒ 1.14 = Mass / 32

⇒ Mass = 36.48 g

Hence, 36.48 g of Oxygen gas is required to react with 40g of Aluminum.