Question

Que cantidad de hidrógeno se necesita para 119gr de amoniaco (NH3) con procedimiento plis

Answer 1

                     3H2 + N2 → 2NH3

Mn NH3 = 17 g/mol
         H2 = 2 g/mol


g H2 = 119 g NH3  x  1 mol NH3  x  3 mol H2  x    2 g H2
                                 ````````````````    ``````````````    ````````````
                                    17 g NH3     2 mol NH3    1 mol H2

g H2 = 21 g necesarios