Question

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

N2 (g) + 3F2(g) → 2NF3 (g)

If a sample containing 19.3 g of N2 is reacted with 19.3 g of F2, how many grams of NF3 will be produced?

Now we need to find the amount of NF₃ that can be formed by the complete reactions of each of the reactants.

If all of the N₂ was used up in the reaction, how many moles of NF₃ would be produced?

Answer 1

N2 (g) + 3F2(g) → 2NF3 (g)

first we should calculate limiting reagent

28 gN2 reacts with 3 x 38 g F2

19.3 g N2 reacts with 19.3 x 3 x 38 / 28 = 78.58 g F2.

but we have only 19.3 g F2

F2 is limiting reagent

3 x 38 g F2 forms 2 x 71 g NF3

19.3 g F2 forms 19.3 x 2 x 71 / 3x 38 = 24.04 g

mass of NF3 formed = 24.04 g

2) 28 g N2 forms 2 x 71 g NF3

19.3 g N2 forms 19.3 x 2 x 71 / 28 = 97.88 g

if complete N2 reacts 97.88 g NF3 will form.

moles of NF3 = 97.88 / 71 = 1.38 moles