Ask Question
8 April, 18:55

How many grams of N2F4 can be produced when 7.00 g of NH3 reacts?

Express your answer with the appropriate units.

+5
Answers (1)
  1. 8 April, 19:17
    0
    21.3 g of dinitrogen tetrafluoride are formed by this reaction

    Explanation:

    First of all, we need to determine the reaction:

    2NH₃ + 5F₂ → N₂F₄ + 6HF

    2 moles of ammonia react with 5 moles of F₂ in order to make 1 mol of dinitrogen tetrafluoride and hydrogen fluoride

    As we only have the ammonia's mass, then the fluorine gas is the excess reagent.

    We convert ammonia's mass to moles → 7 g / 17g/mol = 0.411 moles

    Ratio is 2:1. 2 moles of ammonia produce 1 mol of tetrafluoride

    Therefore, 0.411 moles will produce (0.411. 1) / 2 = 0.205 moles of tetrafluoride

    Let's convert the mass to moles: 0.205 mol. 104.01 g / 1mol = 21.3 g
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “How many grams of N2F4 can be produced when 7.00 g of NH3 reacts? Express your answer with the appropriate units. ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers