Question

4NH3+6NO --> 5N2+6H2O
Si partiendo de 68g de NH3 se obtiene 68g de H2O
¿cuantos gramos de N2 se obtendrán?

Answer 1

            4NH3  + 6NO → 5N2 + 6H2O

Mm NH3 = 17 g/mol
       H2O = 18 g/mol
          N2 = 28 g/mol

1. calcular moles a partir de los gramos        n = masa/Mm

n NH3 = 68 g / 17 g/mol = 4 moles

n H2O = 68 g / 18 g/mol = 3.77 moles

2. calcular moles a partir de la ecuación

n N2 = 4 moles NH3  x   5 moles N2
                                      ``````````````````
                                      4  moles NH3

n N2 = 5 moles

n N2 = 3.77 moles H2O x 5 moles N2
                                         `````````````````
                                        6 moles H2O

n N2 = 3.141 moles

Reactivo limitante: H2O (produce  3.141 moles de N2 )

moles = masa/Mm

masa = 3.141 g x 28 g/mol

masa = 87.94 g de N2