Answer to Question #92878 in General Chemistry for Joshua Tranquilo

Question

If you have 66.7g of NH3, how many grams of F2 are required for complete reaction

Answer 1

Ammonia reacts with flourine in two different ways.

First,$3F_2+2NH_3\to N_2+6HF$

But this reaction is not much recognised.

Second,

It undergoes double displacement reaction to give $N_2F_4$ tetra flouro hydrazine plus hydrogen gas.

$2NH_3+2F_2 \to N_2F_4+3H_2$

I shall solve the question as per this reaction.

As per stoichiometry,

2 moles of NH₃ requires 2 moles of F_2.

Hence,1 mole of $NH_3$ require 1 mole of $F_2$ .

or,17 gm $NH_3 \to 19 gm \space F_2$ (1 mole F₂ =19 gm)

or $66.7 g\space NH_3 \to \frac{19}{17}\times 66.7 \space gm=74.547 \space gm .$(Answer)