1 answer

4 NH₃ + 3 0₂ 2 N2 + 6 H₂O 31 0-26 0 + 36 5....

Question:

4 NH₃ + 3 0₂ 2 N2 + 6 H₂O 31 0-26 0 + 36 5. If 38.7 grams of NH3 react with excess oxygen, how many grams of nitrogen are pro
4 NH₃ + 3 0₂ 2 N2 + 6 H₂O 31 0-26 0 + 36 5. If 38.7 grams of NH3 react with excess oxygen, how many grams of nitrogen are produce (based on the balanced equation in question 4) (6pts) Smoler NH₂ moss/molor mass = 38 2117=2.276 moles moles af No = 1/2*mole sof NH3 = 1/2 * 2.276= 1.138 molos 1,138*28= 31,8649 6. If only 17.9 grams of nitrogen are produced, what is the percent yield (based on your response to question 5)? (3 pts)

Answers

To balance a chemical equation count the number of atoms of each element on the both sides of a reaction and equalise them by adding suitable coefficients.

Balanced equation for this reaction is,

4NH3 + 3O2 ------------> 2N2 + 6H2O

4 moles of NH3 gives 2 moles of N2.

1 mole of NH3 will give 2/4 moles of N2 i.e. 0.5 moles of N2.

Number of moles = Mass/molar mass

Molar mass of NH3 =17.03 g/mol

Moles of NH3 = 38.7/17.03

= 2.27 moles

Moles of N2 formed = 2.27 × 0.5

= 1.14 moles N2.

Molar mass of N2 = 28.01 g/mol

Mass of N2 formed = 28.01 g/mol × 1.14 mol

= 31.9 grams.

Theoretical yield = 31.9 grams.

6. Percent yield = (actual yield/theoretical yield) × 100

Percent yield = (17.9/31.9)×100

= 56.11 %

Percent yield = 56.11 %

.

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