4 NH3 (g) + 5 O2 (g)Â Â -------------> 4 NO (g) + 6
H2O (g)
68
g            Â
160
g                    Â
120
g          Â
108 g
72.0
g          Â
72.0
g                                        Â
??
here
68 g NH3 required ----------------> 160 g O2
72 g NH3 required
------------->Â Â Â Â Â Â Â Â Â
?? O2
mass of O2 = 72 x 160 / 68 = 169.4 g
but here we have only 72.0 g . so
limiting reagent is O2. so product formed according to that.
160 g O2 gives ----------------> 108 g H2O
72.0 g O2 -------------------->Â Â ?? H2O
mass of H2O produced = 72 x 108 / 160 = 48.6 g
maximum mass of H2O that can be produced = 48.6
g
