the balanced equation for the above reaction is as follows;
N₂ + 3I₂ --> 2NI₃
stoichiometry of I₂ to NI₃ is 3:2
mass of NI₃ formed - 3.58 g
The number of NI₃ moles formed - 3.58 g / 394.71 g/mol = 0.00907 mol
according to molar ratio,
when 2 mol of NI₃ is formed - 3 mol of I₂ reacts
therefore when 0.00907 mol is formed - 3/2 x 0.00907 mol = 0.0136 mol of I₂
mass of I₂ reacted - 0.0136 mol x 253.80 g/mol = 3.45 g
therefore mass of I₂ formed - 3.45 g
