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

percent yield = 56.6 %

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 C₂H₄Cl₂ = 2(12) + 4(1) + 2(35.5)

molar mass of C₂H₄Cl₂ = 24 + 4 + 71

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 %

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