Respuesta :

Rmm of N20 = (14X2) + 16 = 44
Percentage by mass of Nitrogen in N20 = 28/44x100 = 63.64
66x0.6634 = 43.7844g of Nitrogen in 66g of N20
Moles = Mass/Mr = 43.7844/14 = 3.13 moles