a) 60 g de funingine=60 g C
60g x
C + O₂ -> CO₂
12g 22,4L
x=(60×22,4)÷12=112L CO₂
b)60g carbune, 10% impuritati=60g C de puritate 90%
[tex] \frac{p}{100} = \frac{mpur}{mimpur} [/tex]
[tex] \frac{90}{100} = \frac{mpur}{60}[/tex] => mpur=(60×90)÷100=54g C
54g x
C + O₂ -> CO₂
12g 22,4L
x=(54×22,4)÷12=100,8L CO₂