The NH3 needed to react exactly with 21 grams of CH3OH is calculated as follows
2NH3 +CH3OH = products
find the moles of CH3OH used
moles = mass/molar mass
= 21 grams/32 g/mol = 0.656 moles
from the mole ratio between NH3 to CH3OH which is 2:1 the moles of NH3 is therefore = 0.656 x2/1 = 1.312 moles
mass of NH3 is therefore= moles of NH3 x molar mass of NH3
= 1.312 moles x 17 g/mol= 22.3 grams