raport molar de 7:1 => 7x moli nitronaftalina si x moli dinitronaftalina
60% din 945=60*945/100=567kg HNO₃
n=m/M=567/63=9kmoli HNO₃
1kmol 1kmol
C₁₀H₈ + HNO₃ -> C₁₀H₇-NO₂ + H₂O
7x 7x
2kmoli 1kmol
C₁₀H₈ + 2HNO₃ -> C₁₀H₆(NO₂)₂ + 2H₂O
2x x
9x=9 => x=1kmol
avem 7kmoli de 1-nitronaftalina
MC₁₀H₇-NO₂=120+7+14+32=173kg/kmol
m=n*M=7*173=1211kg 1-nitronaftalina