Question

If you have 45.6 g NH3, how many grams of F2 are required for a complete reaction? Need to show ALL calculation work.

Answer 1

he balanced reaction equation is:

2NH₃ + 3F₂ $\rightarrow$ N₂ + 6HF.

In this reaction, 2 moles of ammonia react with 3 moles of fluorine:

$\frac{n(NH_3)}{2} = \frac{n(F_2)}{3}$ .

The number of the moles of ammonia is its mass $m$ divided by its molar mass $M$ =17.03 g/mol:

$n(NH_3) = \frac{45.6}{17.03} = 2.678$ mol.

Then, the number of the moles of F₂ needed for the complete reaction is:

$n(F_2) = 3\frac{n(NH_3)}{2} = 4.016$ mol.

Finally, the mass of the fluorine is its number of the moles times its molar mass 38.00 g/mol:

$m(F_2) = n·M = 4.016·38.00 = 152.62$ g.

Answer: if you have 45.6 g NH₃, 152.62 grams of F₂ are required for a complete reaction.