NH4Cl -> NH3+HCl
Pe reactie se foloseste doar masa pura, P%/100 = Masa pura/masa totala . Masa pura = 10,7g
Masa molec. NH4Cl = 14+4+35,5= 53.5
Nr moli NH4Cl = 10,7/53.5 = 0.2 g/mol
Fie x = nr moli NH3
0.2 ..................................x
1........................................1
x= 0.2 => masa NH3 = 0.2(14+3) = 3.4g
Fie y = nr moli HCl
0.2 ...................................... y
1.............................................1
y= 0.2 g/mol