Question

5 F2 (g) + 2 NH3 (g) → N2F4 (g) + 6 HF (g)If you reacted 0.15 moles of F2 how many moles of HF would be produced?

Answer 1

Step 1 - Finding the stoichiometry of the reaction

The given reaction is:

$5F_{2(g)}+2NH_{3(g)}\to N_2F_{4(g)}+6HF_{(g)}$

The stoichiometry of the reaction can be found by the bigger numbers that come before the formula of the substance:

5 moles of F2 react with 2 moles of NH3 thus producing 1 mole of N2F4 and 6 moles of HF

Since the exercise is specifically asking about the relation between F2 and HF, we can simplify this statement to:

5 moles of F2 produce 6 moles of HF

Note that this is a fixed relation, and we'll be using it to solve the question.

Step 2 - Using the stoichiometry to discover how many moles of HF would be produced

We can now set the following proportion, since we already know the stoichiometry:

$\begin{gathered} 5\text{ moles of F2 produce ---- 6 moles of HF} \\ 0.15\text{ moles of F2 would produce --- x} \\ \\ x=\frac{6\times0.15}{5}=0.18\text{ mole of HF} \end{gathered}$