Question

5f2 (g) + 2nh3 (g) → n2f4 (g) + 6hf (g) if you have 58.5g nh3, how many grams of f2 are required for a complete reaction? how many grams of nh3 are required to produce 3.89g hf? how many grams of n2f4 can be produced from 217g f2?

Answer 1

5F2 +2NH3 = N2f4 + 6HF

moles = mass/molar mass
mass =molar mass x mole

a)   moles of Nh3 = 58.5/17 = 3.44 moles
the mole  ratio of F2:NH3 = 5:2
the moles of F2 =3.44x5/2=8.6 moles
mass= 8.6 x 38 = 326.8 grams  of F2

B) moles of Hf = 3.89/20 = 0.1945 moles
mole ratio NH3:Hf =  2:6
moles of  NH3 = 0.1945 x2/6 = 0.0648 moles
mass =0.0648 x17 = 1.102 grams of NH3

C) moles of f2 = 217/38 =5.711 moles
mole ratio N2F6:F2 = 5:1
moles of N2F4 =5.711 x1/5 =1.142 moles
mass = 1.142 x104 = 118.77 grams of N2F4