Question

Carbon reacts with chlorine form ccl4 36gm of carbon was mixed with 142 g of cl2 . calculate mass of ccl4 produced and the remaining mass of reactant.

Answer 1

Reaction for the above is ⇒

C + 2Cl₂ ----------→ CCl₄

In this Reaction, Chlorine is the limiting reactant. We can prove that.

12 g of the carbon reacts with 4 × 35.5 g of the Chlorine.
∴ 1 g of the Carbon reacts with the 142/12 g of the chlorine.
∴ 36 g of the Carbon reacts with the 426 g of the Chlorine.

Since, the mas of the chlorine given is less than the chlorine required. 
∴ Chlorine is the limiting reactant.Thus, whole reaction will be controlled by the Chlorine.

∵ 4 × 35.5 g of the Chlorine reacts with the 12 g of the Carbon.
∴ 1 g of the Chlorine reacts with the 12/142 g of the Carbon.
∴ 142 g of the Chlorine reacts with the 12 g of the Carbon.

Now, the mass of the carbon reacts is less than the mass of the Carbon present, therefore, it is in excess amount.

Now, 
∵ 2 × 35.5 g of the Chlorine produces 154 g of the CCl₄.
∴ 1 g of the chlorine produces 154/142 g of the CCl₄.
∴ 142 g of the CCl₄ produces 154 g of the CCl₄.


Mass of the CCl₄ Produced is 154 g.
Mass of the Carbon left = 36 - 12
 = 24 g.
Mass of the chlorine left = 0 g
[Since, it is completely used up in the reaction].


Hope it helps.

Answer 2

Answer:

Explanation:

Reaction for the above is ⇒

C + 2Cl₂ ----------→ CCl₄

In this Reaction, Chlorine is the limiting reactant. We can prove that.

12 g of the carbon reacts with 4 × 35.5 g of the Chlorine.

∴ 1 g of the Carbon reacts with the 142/12 g of the chlorine.

∴ 36 g of the Carbon reacts with the 426 g of the Chlorine.

Since, the mas of the chlorine given is less than the chlorine required. 

∴ Chlorine is the limiting reactant.Thus, whole reaction will be controlled by the Chlorine.

∵ 4 × 35.5 g of the Chlorine reacts with the 12 g of the Carbon.

∴ 1 g of the Chlorine reacts with the 12/142 g of the Carbon.

∴ 142 g of the Chlorine reacts with the 12 g of the Carbon.

Now, the mass of the carbon reacts is less than the mass of the Carbon present, therefore, it is in excess amount.

Now, 

∵ 2 × 35.5 g of the Chlorine produces 154 g of the CCl₄.

∴ 1 g of the chlorine produces 154/142 g of the CCl₄.

∴ 142 g of the CCl₄ produces 154 g of the CCl₄.

Mass of the CCl₄ Produced is 154 g.

Mass of the Carbon left = 36 - 12

 = 24 g.

Mass of the chlorine left = 0 g

[Since, it is completely used up in the reaction].