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C2H4+

O2
CO2 + H2O

A. What is my theoretical yield of H2O if I start with 9.35 g of Oz?

B. If I actually produced 2.63 g of H20, what is my percent yield?

User Amogh
by
4.2k points

1 Answer

4 votes

Answer:

A. Theoretical yield = 3.51g

B. %yield = 75%

Step-by-step explanation:

The balanced equation for the reaction is given below:

C2H4 + 3O2 —> 2CO2 + 2H2O

Molar Mass of O2 = 16 x 2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96

Molar Mass of H20 = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

A. From the equation,

96g of O2 produced 36g of H2O

Therefore, 9.35g of O2 will produce = (9.35 x 36)/96 = 3.51g of H2O

Therefore,theoretical yield of water (H2O) = 3.51g

B. Theoretical yield = 3.51

Actual yield = 2.63g

%yield =?

%yield = Actual yield/Theoretical yield x 100

%yield = 2.63/3.51

%yield = 75%

User Akash Kumar Verma
by
5.1k points