Answer :
Answer:
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%
Answer:
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 = [tex]\frac{experimental mass}{theoretical mass} x 100[/tex]
% yield = [tex]\frac{280 g}{495 g} x 100[/tex]
% yield = 56.6