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                  

fire could start, buildings collapsing, and ash falling from the sky.

explanation:

this is because the buildings or plants could catch on fire and the buildings would most likely collapse. lastly ash would fall because all the burning would cause ash to fall from the sky.

hope this .

i'm pretty sure it is letter c.) car