Question

cuantos gramos de n2f4 se pueden preparar a partir de 4 gramos de amoniaco y 14 gramos de fluor molecular

Answer 1

  2NH3 + 5F2 → N2F4 + 6HF

Mm N2F4 = 104 g/mol
           F2 = 38 g/mol

1. calcular moles a partir de los gramos

n (NH3) = 4 g /17 g/mol = 0.24 mol

n (F2) =  14 g / 38 g/mol = 0.36 mol

2. calcular moles a partir de la ecuacion

n (N2F4) = 0.24 mol NH3  x  1 mol N2F4
                                              `````````````````
                                               2 mol NH3
n (N2F4) = 0.12 mol

n (N2F4) = 0.36 mol F2  x  1 mol N2F4
                                           `````````````````
                                             5 mol F2
n (N2F4) = 0.072 mol

REACTIVO LIMITANTE: F2 (produce 0.072 mol)

3. calcular masa de N2F4 APLICANDO LA FÓRMULA n = masa/Mm

masa = 0.072 mol x 104 g/mol

masa = 7.49g de N2F4