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?

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