Ce volum ocupa in conditii normale a) 3g H2
b) 12.046x10 la puterea 25 molecule NH3

a) p=1atm

T=273°K

R=0.082 atm/mol×K

P×V=niuRT

n=m/M

M, H2=1×2=2g/mol

n=3/2=1.5 moli

V=niuRT/P=1.5×0.082×273/1=33.57L

b) 1mol NH3.........6.023×10^23molec NH3

X........................12.046×10^25molec

X=12.046×10^25/6.023×10^23=200 moli NH3

1mol.............22.4l

200moli......y

Y=200×22.4=4480 l