ms = 635 kg, c% = 20%
c% = mdx100/ms
=> md = msxc%/100
= 635x20/100 = 127 kg HNO3
notam cu a = masa de HNO3 consumata din cele 127 kg
c.f% = 2%
md.f = md - a
ms.f = ms - a
=> c.f% = md.fx100/ms.f
=> 2 = (127-a)100/(635-a)
=> 1270 - 2a = 12700 - 100a
=> a = 11430/98 = 116,63 kg HNO3 consumat
116,63 kg m kg
C6H6 + HNO3 --> C6H5NO2 + H2O
63 123
=> m = 227,71 kg
=> C)