Trebuie sa afli cantitatea de H continuta in 12 moli apa 1mol H2O contine 2 g H 12 moli contin X X=12x2=24 gH MNH3=14+3=17g/mol 17 gNH3 contin 3 g H Yg contin 24g H Y=24x17/3=136 gNH3 M CH4=12+4=16g/mol 16g CH4 contin 4g H Z g contin 24 gH Z=24x16/4=96 gCH4
