calculez moli N2
n=V/Vm= 67,2 l/22,4 l/mol= 3mol N2
din ecuatia reactiei, deduc moli NH3 obtinuti din 3 moli N2
1mol..............2mol
N2 + 3H2---> 2NH3
3mol....................x= 6molNH3
APLIC ecuatia de stare a gazelor
pV= nRT
4,1XV=6x0,082X(273+27),,,,,,,,,,,,,,,,,,,,,,,CALCULAEZA V !!!!!