Molar mass
NH3 = 17 g/mol
N2 = 28 g/mol
4 NH3 + 6 NO = 5 N2 + 6 H2O
4 x 17 g NH3 ------------ 5 x 28 g N2
?? g NH3 --------------- 300 g N2
300 x 4 x 17 / 5 x 28 =
20400 / 140 => 145.71 g of NH3
An estuary is partly enclosed and forms a transition zone between rivers and oceans/ seas
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
