When ammonia (NH_3) reacts with fluorine, the products are gaseous dinitrogen te

Question

When ammonia (NH_3) reacts with fluorine, the products are gaseous dinitrogen tetrafluoride (N_2F_4) and hydrogen fluoride (HF). Express your answer as a chemical equation. Identify all of the phases in your answer. 2NH_3(g) + 5F_2 (g) rightarrow N_2F_4 (g) + 6HF(g) How many moles of each reactant are needed to produce 1.91 moles of HF? Enter your answers numerically separated by a comma. How many grams of F_2 are required to react with 35.9 g of NH_3? Express your answer with the appropriate units. How many grams of N_2F_4 can be produced when 3.83 g of NH_3 reacts? Express your answer with the appropriate units.

Explanation / Answer

C)

mass of NH3 = 35.9 g

molar mass of NH3 = 17.03 g/mol

mol of NH3 = (mass)/(molar mass)

= 35.9/17.03

= 2.11 mol

According to balanced equation

mol of F2 reacted = (5/2)* moles of NH3

= (5/2)*2.11

= 5.27 mol

masss of F2 = number of mol * molar mass

= 5.27*38

= 200 g

Answer: 200 g

D)

mass of NH3 = 3.83 g

molar mass of NH3 = 17.03 g/mol

mol of NH3 = (mass)/(molar mass)

= 3.83/17.03

= 0.225 mol

According to balanced equation

mol of N2F4 formed = (1/2)* moles of NH3

= (1/2)*0.225

= 0.1125 mol

masss of N2F4 = number of mol * molar mass

= 0.1125*104

= 11.7 g

Answer: 11.7 g

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