Question

A chemical engineer calculated that 15.0 mol H2 was needed to react with excess N2 to prepare 10.0 mol NH3. But the percentage yield of the reaction is 60.0%.

a. Write a balanced equation for the reaction
b. How many moles of NH3 were actually made based on the percentage yield?
c. How many grams of NH3 are formed?

Answer 1

A. 3H2 + N2 —> 2NH3

B. 6 moles

C. 170g

Step-by-step explanation:

A. The balanced equation for the reaction? This is given below:

3H2 + N2 —> 2NH3

B. Determination of the actual yield of NH3. This is illustrated below:

Theoretical yield = 10 moles

Percentage yield = 60%

Actual yield =?

Percentage yield = Actual yield/Theoretical yield x100

60/100 = Actual yield /10

0.6 = Actual yield/10

Cross multiply

Actual yield = 0.6 x 10

Actual yield of NH3 = 6 moles

C. Determine the mass of NH3 produced.

Number of mole NH3 = 10 moles

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 =?

Mass = number of mole x molar Mass

Mass of NH3 = 10 x 17

Mass of NH3 = 170g