75.0g ammonia NH3 reacts with 197.0 g fluorine to produce dinitrogen tetrafluoride and hydrogen fluoride (used in production of aluminum, in uranium processing, and in frosting of light bulbs) according to the unbalanced equation below.                       NH3(g) + F2(g) → N2F4(g) + HF(g) Answer all the following questions, show all the work and calculations. Pay attention to significant figures: Which reagent is the limiting reagent? How many moles of N2F4(g) is produced during this reaction? What is the theoretical yield in grams of N2F4(g) Given that actual yield of N2F4(g) is 94.0 g, what is the percent yield of N2F4(g)?

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Please solve last two sub-parts......3,4

75.0g ammonia NH3 reacts with 197.0 g fluorine to produce dinitrogen tetrafluoride and hydrogen fluoride (used in production of aluminum, in uranium processing, and in frosting of light bulbs) according to the unbalanced equation below.

 

                    NH3(g) + F2(g) → N2F4(g) + HF(g)

Answer all the following questions, show all the work and calculations. Pay attention to significant figures:

  1. Which reagent is the limiting reagent?
  2. How many moles of N2F4(g) is produced during this reaction?
  3. What is the theoretical yield in grams of N2F4(g)
  4. Given that actual yield of N2F4(g) is 94.0 g, what is the percent yield of N2F4(g)?
 
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