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

Express your answer with the appropriate units.

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