C 2 H 4 + Cl 2 ---------> C 2 H 4 Cl 2 In the above reaction; C 2 H 4 and Cl 2 r

Question

C2H4     +     Cl2       --------->       C2H4Cl2

In the above reaction; C2H4 and Cl2 reacts in 1:1 ratio

11.22 g of C2H4
Molar mass of C2H4 = 28.05 g/mol
So, 28.05 g of C2H4 = 1 mol
or, 1 g of C2H4 = (1/28.05) mol
or, 11.22 g of C2H4 = (11.22/28.05) mol = 0.4 mol

35.45 g of Cl2
Molar mass of Cl2 = 70.90 g/mol
So, 70.90 g of Cl2 = 1 mol
or, 1 g of Cl2 = (1/70.90) mol
or, 35.45 g of Cl2 = (35.45 / 70.90) mol = 0.5 mol

Since, Cl2 is used as excess.
So, C2H4 is limiting reagent.

Now,

1 mole of C2H4 reacts with 1 mole of Cl2 to produce 1 mole of C2H4Cl2.
or, 0.4 mole of C2H4 reacts with 0.4 mole of Cl2 to produce 0.4 mole of C2H4Cl2.

Molar mass of C2H4Cl2 = 98.95 g/mol
So, 1 mole of C2H4Cl2 = 98.95 g
or, 0.4 mole of C2H4Cl2 = 0.4 x 98.95 g = 39.58 g

39.58 g is the theoretical yield or (100% yield).

(b)

50% yield

So, 50% of 0.4 moles = 0.2 moles

Moles remained unreacted are
C2H4 = 0.4 mol – 0.2 mol = 0.2mol
Cl2 = 0.5 mol – 0.2 mol = 0.3 mol

So, total mass remained unreacted is

(0.2 mol x 28.05 g/mol) + (0.3 mol x 70.90 g/mol)
= 26.88 g

Explanation / Answer

Page % of ,X) ( 10 pre o points) The dry-cleaning solvent 1,2 dichloroethane (also called ethylene chloride) CHACt. is The by reaction of 11.22 g of ethylene (CHs) with 35.45 g of Cl C2H4+Cl2CHAC12 (a) If the percent yield is 100.0%, after the reaction, how many grams of the product of 1,2 produced? (b f the percent yield is 50.00%, after the reaction, what is the total mass of the rea tan set oven 1.2 dichloroethane ar mass of C2H4-28.05 g/mol; Molar mass of C12 = 70.90 g/mol Molar mass of C2H C½=985% mo

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