El amoniaco (NH3) se descompone en dinitrógeno (N2) y dihidrógeno (H2) gaseosos?

(a) 2 NH3 → 3H2 + N2

(b) Si se obtienen 28 g de dinitrógeno y 6 g dihidrógeno, ¿cuánto amoniaco es necesario?

Mm N2 = 28 g / mol NH3 = 17 g / mol

calcular moles de N2 y de H2

mol N2 = 28 g x 1 mol N2 ```````````` 28 g N2

mol N2 = 1

mol H2 = 6 g H2 x 1 mol H2 `````````````` 2 g H2

mol H2 = 3

Calcular moles de NH3 a partir de los moles de calculados

mol NH3 = 1 mol N2 x 2 mol NH3 ```````````````` 1 mol N2

mol NH3 = 2

mol NH3 = 3 mol H2 x 2 mol NH3 ```````````````` 3 mol H2

mol NH3 = 2

g NH3 = 17 g / mol x21 mol = 34 g de NH3

calcular gramos de N2

c) g N2 = 17 g NH3 x 1 mol NH3 x 1 mol N2 x 28 g N2 `````````````` ````````````` `````````````` 17 g NH3 2 mol NH3 1 mol N2

g N2 = 14 g

Calcular gramos de hidrógeno

g H2 = 17 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 = 3 g

(c) A partir de 17 g de amoniaco, ¿cuánto nitrógeno e hidrógeno se desprende?