Question

What is the % yield when 140.0 grams of Ethylene gas (C2H4) reacts with excess chlorine to form 280.0 grams of 1,2-Dichloro Ethane (C2H4Cl2)? (Given equation is C2H4 + Cl2 —> C2H4Cl2

Answer 1

percent yield = 56.6 %

Step-by-step explanation:

Data given:

mass of Ethylene gas (C₂H₄) = 140 g

actual yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 280 g

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = ?

Reaction Given:

C₂H₄ + Cl₂ -------> C₂H₄Cl₂

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

1 mol 1 mol

As 1 mole of C give 1 mole of CH₄

Convert moles to mass

molar mass of C₂H₄ = 2(12) + 4(1)

molar mass of C₂H₄ = 24 + 4

• molar mass of C₂H₄ = 28 g/mol

molar mass of C₂H₄Cl₂ = 2(12) + 4(1) + 2(35.5)

molar mass of C₂H₄Cl₂ = 24 + 4 + 71

• molar mass of C₂H₄Cl₂ = 99 g/mol

Now

C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

1 mol (28 g/mol) 1 mol (99 g/mol)

28 g 99 g

28 grams of Ethylene gas (C₂H₄) produce 99 grams of C₂H₄Cl₂

So

if 28 g of C₂H₄ produce 99 g of C₂H₄Cl₂ so how many grams of C₂H₄Cl₂ will be produced by 140 g of C₂H₄.

Apply Unity Formula

28 g of C₂H₄ ≅ 99 g of C₂H₄Cl₂

140 g of C₂H₄≅ X of C₂H₄Cl₂

Do cross multiply

mass of C₂H₄Cl₂ = 99 g x 140 g / 28 g

mass of C₂H₄Cl₂ = 495 g

So the Theoretical yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 495 g

Now Find the percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

percent yield = 280g / 495 g x 100 %

percent yield = 56.6 %

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = 56.6 %