Respuesta :

kenmyna
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