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

The answer to your question is percent yield = 56.6

Step-by-step explanation:

Balanced Reaction

C₂H₄ + Cl₂ ⇒ C₂H₄Cl₂

Data

Reactants

C₂H₄ 140 g

Cl₂ excess

Products

C₂H₄Cl₂ 280 g

Process

1.- Calculate the molecular mass of Ethylene and Dichloro ethane.

Ethylene = (12 x 2) + (1 x 4) = 24 + 4 = 28 g

Dichloro ethane = (12 x 2) + (1 x 4) + (35.5 x 2) = 24 + 4 + 71 = 99 g

2.- Calculate the theoretical amount of dichloro ethane using proportions.

28 g of Ethylene --------------- 99 g of Dichloro ethane

140 g of Ethylene --------------- x

x = (140 g x 99 g) / 28 g

x = 495 g of Dichloro ethane

3.- Calculate the percent yield

% yield =
`(experimental mass)/(theoretical mass) x 100`

% yield =
`(280 g)/(495 g) x 100`

% yield = 56.6

Answer 2

The % yield is 56.6 %

Step-by-step explanation:

This is the reaction:

C₂H₄ + Cl₂ → C₂H₄Cl₂

Molar mass of ethylene gas: 28 g/m

Mol = mass / molar mass

140 g / 28 g/m = 5 moles

Ratio is 1:1, so 5 moles of ethylene produce 5 moles of dichloro ethane.

Molar mass of C₂H₄Cl₂ = 98.9 g/m

Mass of C₂H₄Cl₂ produced = 98.9 g/m . 5 m → 494.5 g

% yield reaction

(280 g / 494.5 g ) . 100 = 56.6%