If you have 66.6 g NH3, how many grams of F2 are needed to consume all the NH3?

The balanced equation for this reaction is:

  `~2NH_3 ` + `~5F_2` -> `~N_2F_4 ` + `~6HF`

Step 1: Determine the molar mass of `~NH_3` and `~F_2` .

Since "grams" are involved in this calculation, we will need to use the following mole conversion factors: 1 mole = molar mass `~NH_3` AND 1 mole = molar mass `~F_2` . So, before we start the main calculation, let's go ahead and determine the molar mass of `~NH_3` and `~F_2` .

The molar mass of a substance is determined by multiplying the atomic mass of each atom in the substance times its subscript and adding the resulting answers.

   Molar mass of `~NH_3` = (1)(14.007) + 3(1.008) = 17.031 g/mol

   Molar mass of `~F_2` = (2)(18.998) = 37.996 g/mol

Therefore,

   1 mole `~NH_3` = 17.031 grams

   1 mole `~F_2` = 37.996 grams

Step 2: Determine the mole ratio between `~NH_3` and `~F_2` .

The mole ratio between two substances is equal to the ratio between the coefficients of the substances. According the reaction above, the coefficient for `~NH_3 ` is 2 and the coefficient for `~F_2` is 5. 

Therefore, the mole ratio between `~NH_3` and `~F_2` is:

   2 moles `~NH_3 ` = 5 moles `~F_2`

Step 3: Perform the stoichiometry calculation.

The stoichiometry calculation will take the general form of:

  given amount  x  `~NH_3` mole conversion factor  x  mole ratio  x  `~F_2` mole conversion factor

Therefore, 

   66.6 g `~NH_3` x (1 mol/17.031) x (5 mol `~F_2` /2 mol `~NH_3` ) x (37.996 g/1mol) = 371 g `~F_2`

*Notice that the conversion factors and mole ratio are oriented such that all units and substances cancel out except grams of `~F_2`.