M(NH4NO3)= 14+1*4+14+16*3 = 80 g/mol
W(N) = (14+14) / 80 * 100% = 35%
ответ 35% азота
M(KNO3) = 39+14+48 = 101 g/mol
W(N)= 14 / 101 * 100% = 13.86%
M(K2O) = 39*2+16 = 94 g/mol
W(K2O) = 94 / 101 *100% = 93.07%
ответ 13.86 % азота , 93.07% - оксида калия