Question

sulphur and oxygen with atomic masses 32u and 16u respectively react to form Sulphur trioxide. How much Sulphur will react with 3g oxygen to form Sulphur trioxide -


4g

2g

3g

32g​

Answer 1

Answer: 2g

Given is the following reaction according to the question,

S + O₂ ➞ SO₃

First let's balance the given reaction,

2S + 3O₂ ➞ 2SO₃

Also, Atomic mass of:

• Sulphur, S = 32u
• Oxygen, O = 16u

Now, We have to find how much Sulphur will react to 3g to form Sulphur Trioxide.

From the reaction, It is clear that 2 moles of Sulphur reacts with 3 moles of O₂.

It can also be expressed as,

⇒ 2 moles S ➞ 3 moles O₂

So, 1 mole of O₂ will react with 2/3 mole of S.

So, Let's find the moles of O₂, given

• Molar mass = 16×2 = 32g
• Mass = 3g

∴ Mole = Mass / Molar mass

Hence, Mole = 3/32 mole

So, Let's find the moles of S then,

⇒ 1 mole O₂ = 2/3 mole S

⇒ 3/32 mole = 2/3 × 3/32 mole S

⇒ 3/32 mole = 1/16 mole S

Hence, 1/16 mole of S will react with 3g of O₂.

Let's find the mass now,

⇒ Moles = Mass / Molar mass

⇒ 1/16 = mass / 32

⇒ 32/16 = mass

⇒ mass = 2g

Hence, 2g of Sulphur will react with 3g of O₂.

Answer 2

Answer :-

At first finding Oxygen

$\sf Molar \; mass = Mass \; of \; oxygen \times2$

$\sf Molar \; Mass = 16\times 2$

$\sf Molar \; mass = 32 g$

$\sf Mass = 3g$

As we are knowing that

$\sf Mole = \dfrac{Mass}{Molar \; mass}$

Mole = 3/32

Finding Sulphur

$\tt 1 \; mole = \dfrac{2}{3} \times Mole \;of \;o$

$\tt 1 \; mole = \dfrac{2}3 \times \dfrac{3}{32}$

$\tt 1 \; mole = \dfrac{2}{32}$

$\bf 1 \; mole = \dfrac{1}{13}$

Finding mass

$\sf Mass = Moles \times Molar \; mass$

$\tt Mass = \dfrac{1}{16} \times 32$

$\tt Mass = 1 \times 2$

$\bf Mass =2 \; g \bigg(Option \; B \bigg)$