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
