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2 years ago

How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:

Chemistry

1 answer:

WARRIOR [948]2 years ago

8 0

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol 3 mol 1 mol 1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 + 2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

1 mol (76 g/mol) 3 mol (71 g/mol)

CS₂ + 3Cl₂ --------> CCl₄ + S₂Cl₂

76 g 213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

CS₂ : Cl₂

76 g : 213 g

1 g : 2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

2.8 g of Cl₂ ≅ 1 g of CS₂

10 g of Cl₂ ≅ ? g of CS₂

by doing cross multiplication

g of CS₂ = 10 x 1 / 2.8

g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

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