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

The % yield is 56.6 %

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%

The answer to your question is percent yield = 56.6

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 =

% yield =

% yield = 56.6                  

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