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 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)

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